Chapter 10 Conic Sections

NCERT Solutions Class 11 Maths Chapter 10 – Conic Sections

NCERT Solutions for Class 11 Maths Chapter 10 (Conic Sections) explore the curves obtained by the intersection of a plane with a double-napped right circular cone. This chapter transitions from the study of straight lines to more complex curves: Circles, Parabolas, Ellipses, and Hyperbolas. These shapes are not just mathematical abstractions; they describe the orbits of planets, the path of a projectile, and the design of satellite dishes and bridges.

The NCERT Solutions for Class 11 Maths Chapter 10 by ALLEN are designed by expert faculty to simplify these geometric curves through clear derivations of their standard equations. The solutions focus on the relationship between the focus, directrix, and eccentricity, providing a systematic approach to identifying and sketching each conic section.

Mastering Conic Sections is vital for success in JEE Main and JEE Advanced, as this chapter forms a significant portion of the Coordinate Geometry section. These solutions provide the rigorous logical framework needed to find the length of the latus rectum, coordinates of the foci, and equations of the directrices, which are frequently tested in engineering entrance exams.

1.0Class 11 Maths Chapter 10 : Key Concepts

This chapter focuses on defining each curve as a locus of points and deriving their standard mathematical equations. Key lessons include:

Sections of a Cone: Understanding how varying the angle of the intersecting plane results in different curves.
Circle: The set of all points in a plane that are equidistant from a fixed point (center).
Standard Equation: $(x - h)^{2} + (y - k)^{2} = r^{2} .$
Parabola: The set of points equidistant from a fixed line (directrix) and a fixed point (focus) is called a parabola.
Standard Forms: $y^{2} = 4 a x, y^{2} = - 4 a x, x^{2} = 4 a y, x^{2} = - 4 a y .$
Ellipse: The set of points where the sum of distances from two fixed points (foci) is constant is called an ellipse.
Standard Equation: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1.$
Key components: Major axis, Minor axis, Eccentricity (e < 1), and Latus Rectum.
Hyperbola: The set of points where the absolute difference of distances from two fixed points (foci) is constant.
Standard Equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1.$
Eccentricity (e > 1) and Transverse/Conjugate axes.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 9

EXERCISE - 10.1

In each of the following Q. 1 to 5, find the equation of the circle with

Centre $(0, 2)$ and radius 2. Sol. The equation of a circle with centre (h, k) and radius $r$ is given as $(x - h)^{2} + (y - k)^{2} = r^{2}$ Here, therefore the equation of the circle is $(x - 0)^{2} + (y - 2)^{2} = 2^{2}$ $\Rightarrow x^{2} + y^{2} + 4 - 4 y = 4$ $\Rightarrow x^{2} + y^{2} - 4 y = 0$
Centre (-2, 3) and radius 4. Sol. The equation of a circle with centre (h, k) and radius $r$ is given as $(x - h)^{2} + (y - k)^{2} = r^{2}$ Here, therefore the equation of the circle is $(x + 2)^{2} + (y - 3)^{2} = (4)^{2}$ $\Rightarrow x^{2} + 4 x + 4 + y^{2} - 6 y + 9 = 16$ $\Rightarrow x^{2} + y^{2} + 4 x - 6 y - 3 = 0$
Centre $(\frac{1}{2}, \frac{1}{4})$ and $\frac{1}{12}$ radius. Sol. The equation of a circle with centre (h, k) and radius $r$ is given as $(x - h)^{2} + (y - k)^{2} = r^{2}$ Here, therefore, the equation of the circle is

$(x - \frac{1}{2})^{2} + (y - \frac{1}{4})^{2} = (\frac{1}{12})^{2}$

$\Rightarrow x^{2} - x + \frac{1}{4} + y^{2} - \frac{y}{2} + \frac{1}{16} - \frac{1}{144}$ $\Rightarrow x^{2} - x + \frac{1}{4} + y^{2} - \frac{y}{2} + \frac{1}{16} - \frac{1}{144} = 0$ $\Rightarrow 144 x^{2} - 144 x + 36 + 144 y^{2} - 72 y + 9 - 1 = 0$ $\Rightarrow 144 x^{2} - 144 x + 144 y^{2} - 72 y + 44 = 0$ $\Rightarrow 36 x^{2} - 36 x + 36 y^{2} - 18 y + 11 = 0$

4. Centre $(1, 1)$ and radius $2$ . Sol. The equation of a circle with centre (h, k) and radius $r$ is given as $(x - h)^{2} + (y - k)^{2} = r^{2}$ Here, therefore, the equation of the circle is

$(x - 1)^{2} + (y - 1)^{2} = (2)^{2}$

$\Rightarrow x^{2} - 2 x + 1 + y^{2} - 2 y + 1 = 2$ $\Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0$

5. Centre $(- a, - b)$ and radius $a^{2} - b^{2}$ . Sol. The equation of a circle with centre (h, k) and radius $r$ is given as

$(x - h)^{2} + (y - k)^{2} = r^{2}$

Here, therefore, the equation of the circle is

$(x + a)^{2} + (y + b)^{2} = (a^{2} - b^{2})^{2}$

$\Rightarrow x^{2} + 2 ax + a^{2} + y^{2} + 2 by + b^{2} = a^{2} - b^{2}$ $\Rightarrow x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0$

In each of the following Q. 6 to 9, find the centre and radius of the circles.

$(x + 5)^{2} + (y - 3)^{2} = 36$ Sol. The equation of the given circle is $(x + 5)^{2} + (y - 3)^{2} = 36$ . $\Rightarrow {x - (- 5)}^{2} + (y - 3)^{2} = 6^{2}$

which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $h = - 5, k = 3$ , and $r = 6$ . Thus, the centre of the given circle is (-5, 3), while its radius is 6.

$x^{2} + y^{2} - 4 x - 8 y - 45 = 0$ Sol. The equation of the given circle is $x^{2} + y^{2} - 4 x - 8 y - 45 = 0$ . $\Rightarrow (x^{2} - 4 x) + (y^{2} - 8 y) = 45$ $\Rightarrow {x^{2} - 2 (x) (2) + 2^{2}} + {y^{2} - 2 (y) (4) + 4^{2}} - 4 - 16 = 45$ $\Rightarrow (x - 2)^{2} + (y - 4)^{2} = 65$ $\Rightarrow (x - 2)^{2} + (y - 4)^{2} = 65$

Which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $h = 2, k = 4$ and $r = 65$ , Thus, the centre of the given circle is (2, 4), while its radius is $65$ .

$x^{2} + y^{2} - 8 x + 10 y - 12 = 0$ Sol. The equation of the given circle is $x^{2} + y^{2} - 8 x + 10 y - 12 = 0$ $\Rightarrow (x^{2} - 8 x) + (y^{2} + 10 y) = 12$ $\Rightarrow {x^{2} - 2 (x) (4) + 4^{2}} + {y^{2} + 2 (y) (5) + 5^{2}} - 16 - 25 = 12$ $\Rightarrow (x - 4)^{2} + (y + 5)^{2} = 53$ $(x - 4)^{2} + {y - (- 5)}^{2} = (53)^{2}$

which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $h = 4, k = - 5$ and $r = 53$ Thus, the centre of the given circle is $(4, - 5)$ , while its radius is $53$ .

9. $2 x^{2} + 2 y^{2} - x = 0$ Sol. The equation of the given circle is $2 x^{2} + 2 y^{2} - x = 0$

$(2 x^{2} - x) + 2 y^{2} = 0$

$\Rightarrow 2 [(x^{2} - \frac{x}{2}) + y^{2}] = 0$ $\Rightarrow {x^{2} - 2 x (\frac{1}{4}) + (\frac{1}{4})^{2}} + y^{2} - (\frac{1}{4})^{2} = 0$ $\Rightarrow (x - \frac{1}{4})^{2} + (y - 0)^{2} = (\frac{1}{4})^{2}$ which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $h = \frac{1}{4}, k = 0$ and $r = \frac{1}{4}$ . Thus, the centre of the given circle is $(\frac{1}{4}, 0)$ , while its radius is $\frac{1}{4}$ .

10. Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4 x + y = 16$ . Sol. Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Since the circle passes through points $(4, 1)$ and (6, 5),

$(4 - h)^{2} + (1 - k)^{2} = r^{2} (6 - h)^{2} + (5 - k)^{2} = r^{2}$

Since the centre $(h, k)$ of the circle lies on line $4 x + y = 16$ ,

$4 h + k = 16$

From equations (1) and (2), we obtain

$(4 - h)^{2} + (1 - k)^{2} = (6 - h)^{2} + (5 - k)^{2}$

⇒

$16 - 8 h + h^{2} + 1 - 2 k + k^{2} = 36 - 12 h + h^{2} + 25 - 10 k + k^{2}$

$\Rightarrow 16 - 8 h + 1 - 2 k = 36 - 12 h + 25 - 10 k$ $\Rightarrow 4 h + 8 k = 44$ $\Rightarrow h + 2 k = 11$ On solving equations (3) and (4), we obtain $h = 3$ and $k = 4$ . On substituting the values of h and k in equation (1), we obtain

$(4 - 3)^{2} + (1 - 4)^{2} = r^{2}$

$\Rightarrow (1)^{2} + (- 3)^{2} = r^{2}$ $\Rightarrow 1 + 9 = r^{2}$ $\Rightarrow r^{2} = 10$ $\Rightarrow r = 10$ Thus, the equation of the required circle is

$(x - 3)^{2} + (y - 4)^{2} = (10)^{2}$

$\Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = 10$ $\Rightarrow x^{2} + y^{2} - 6 x - 8 y + 15 = 0$

11. Find the equation of the circle passing through the points $(2, 3)$ and $(- 1, 1)$ and whose centre is on the line $x - 3 y - 11 = 0$ . Sol. Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Since the circle passes through points $(2, 3)$ and (-1, 1),

$(2 - h)^{2} + (3 - k)^{2} = r^{2} (- 1 - h)^{2} + (1 - k)^{2} = r^{2}$

Since the centre $(h, k)$ of the circle lies on line $x - 3 y - 11 = 0$ , $h - 3 k = 11$ From equations (1) and (2), we obtain

$(2 - h)^{2} + (3 - k)^{2} = (- 1 - h)^{2} + (1 - k)^{2} 4 - 4 h + h^{2} + 9 - 6 k + k^{2} = 1 + 2 h + h^{2} + 1 - 2 k + k^{2} 4 - 4 h + 9 - 6 k = 1 + 2 h + 1 - 2 k 6 h + 4 k = 11$

On solving equations (3) and (4), we obtain

$h = \frac{7}{2} and k = \frac{- 5}{2} .$

On substituting the values of h and k in equation (1), we obtain

$(2 - \frac{7}{2})^{2} + (3 + \frac{5}{2})^{2} = r^{2}$

$\Rightarrow (\frac{4 - 7}{2})^{2} + (\frac{6 + 5}{2})^{2} = r^{2}$ $\Rightarrow (\frac{- 3}{2})^{2} + (\frac{11}{2})^{2} = r^{2}$ $\Rightarrow \frac{9}{4} + \frac{121}{4} = r^{2}$ $\Rightarrow \frac{130}{4} = r^{2}$ Thus, the equation of the required circle is

$(x - \frac{7}{2})^{2} + (y + \frac{5}{2})^{2} = \frac{130}{4} (\frac{2 x - 7}{2})^{2} + (\frac{2 y + 5}{2})^{2} = \frac{130}{4} 4 x^{2} - 28 x + 49 + 4 y^{2} + 20 y + 25 = 130 4 x^{2} + 4 y^{2} - 28 x + 20 y - 56 = 0 4 (x^{2} + y^{2} - 7 x + 5 y - 14) = 0 x^{2} + y^{2} - 7 x + 5 y - 14 = 0$

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3). Sol. Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Since the radius of the circle is 5 and its centre lies on the x-axis, $k = 0$ and $r = 5$ . Now, the equation of the circle becomes $(x - h)^{2} + y^{2} = 25$ . It is given that the circle passes through point $(2, 3)$ . $∴ (2 - h)^{2} + 3^{2} = 25$ $\Rightarrow (2 - h)^{2} = 25 - 9$ $\Rightarrow (2 - h)^{2} = 16$ $\Rightarrow (2 - h)^{2} = \pm 16 = \pm 4$ If $2 - h = 4$ , then $h = - 2$ If $2 - h = - 4$ , then $h = 6$ When $h = - 2$ , the equation of the circle becomes $∴ (x + 2)^{2} + y^{2} = 25$ $\Rightarrow x^{2} + 4 x + 4 + y^{2} = 25$ $\Rightarrow x^{2} + y^{2} + 4 x - 21 = 0$ When $h = 6$ , the equation of the circle becomes $∴ (x - 6)^{2} + y^{2} = 25$ $\Rightarrow x^{2} - 12 x + 36 + y^{2} = 25$ $\Rightarrow x^{2} + y^{2} - 12 x + 11 = 0$
Find the equation of the circle passing through $(0, 0)$ and making intercepts a and b on the coordinate axes. Sol. Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Since the centre of the circle passes through (0, 0),

$(0 - h)^{2} + (0 - k)^{2} = r^{2}$

$\Rightarrow h^{2} + k^{2} = r^{2}$ The equation of the circle now becomes $(x - h)^{2} + (y - k)^{2} = h^{2} + k^{2}$ It is given that the circle makes intercepts a and $b$ on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

$(a - h)^{2} + (0 - k)^{2} = h^{2} + k^{2} (0 - h)^{2} + (b - k)^{2} = h^{2} + k^{2}$

From equation (1), we obtain

$a^{2} - 2 ah + h^{2} + k^{2} = h^{2} + k^{2}$

$\Rightarrow a^{2} - 2 ah = 0$ $\Rightarrow a (a - 2 h) = 0$ $\Rightarrow a = 0$ or $(a - 2 h) = 0$ However, $a \neq = 0$ ; hence, $(a - 2 h) = 0$ $\Rightarrow h = \frac{a}{2}$ From equation (2), we obtain

$h^{2} + b^{2} - 2 bk + k^{2} = h^{2} + k^{2}$

$\Rightarrow b^{2} - 2 bk = 0$ $\Rightarrow b (b - 2 k) = 0$ $\Rightarrow b = 0$ or $(b - 2 k) = 0$ However, $b \neq = 0$ ; hence, $(b - 2 k) = 0 \Rightarrow k = \frac{b}{2}$ .

Thus, the equation of the required circle is

$(x - \frac{a}{2})^{2} + (y - \frac{b}{2})^{2} = (\frac{a}{2})^{2} + (\frac{b}{2})^{2}$

$\Rightarrow (\frac{2 x - a}{2})^{2} + (\frac{2 y - b}{2})^{2} = \frac{a ^{2} + b ^{2}}{4}$ $\Rightarrow 4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$ $\Rightarrow 4 x^{2} + 4 y^{2} - 4 ax + 4 by = 0$ $\Rightarrow x^{2} + y^{2} - ax + by = 0$

Find the equation of a circle with centre $(2, 2)$ and passes through the point $(4, 5)$ . Sol. The centre of the circle is given as $(h, k) = (2, 2)$ . Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points $(2, 2)$ and (4, 5).

$∴ r = (2 - 4)^{2} + (2 - 5)^{2} = (- 2)^{2} + (- 3)^{2} = 4 + 9 = 13$

Thus, the equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$ $\Rightarrow (x - 2)^{2} + (y - 2)^{2} = (13)^{2}$ $\Rightarrow x^{2} - 4 x + 4 + y^{2} - 4 y + 4 = 13$ $\Rightarrow x^{2} - y^{2} - 4 x + 4 y - 5 = 0$

Does the point (-2.5, 3.5) lie inside, outside or on the circle $x^{2} + y^{2} = 25$ ? Sol. The equation of the given circle is $x^{2} + y^{2} = 25$ .

$x^{2} + y^{2} = 25 \Rightarrow (x - 0)^{2} + (y - 0)^{2} = 5^{2},$

which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $h = 0, k = 0$ , and $r = 5$ . $∴$ Centre $= (0, 0)$ and radius = 5 Distance between point (-2.5, 3.5) and centre $(0, 0)$

$= (- 2.5 - 0)^{2} + (3.5 - 0)^{2} = 6.25 + 12.25 = 18.5 = 4.3 (approx.) < 5$

Since the distance between point (-2.5, 3.5) and centre $(0, 0)$ of the circle is less than the radius of the circle, point (-2.5, 3.5) lies inside the circle.

EXERCISE - 10.2

In each of the following Q. 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^{2} = 12 x$ .

Sol. The given equation is $y^{2} = 12 x$ . Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with $y^{2} = 4 ax$ , we obtain $4 a = 12 \Rightarrow a = 3$ ∴ Coordinates of the focus $= (a, 0) = (3, 0)$ Since the given equation involves $y^{2}$ , the axis of the parabola is the x-axis. Equation of direcctrix, $x = - a$ i.e., $x = - 3$ i.e., $x + 3 = 0$ Length of latus rectum $= 4 a = 4 \times 3 = 12$

$x^{2} = 6 y$

Sol. The given equation is $x^{2} = 6 y$ . Here, the coefficient of y is positive. Hence, the parabola opens upwards. On comparing this equation with $x^{2} = 4 ay$ , we obtain

$4 a = 6 \Rightarrow a = \frac{3}{2}$

∴ Coordinates of the focus $= (0, a) = (0, \frac{3}{2})$ Since the given equation involves $x^{2}$ , the axis of the parabola is the y-axis. Equation of directrix, $y = - a$ , i.e., $y = - \frac{3}{2}$ Length of latus rectum $= 4 a = 6$

$y^{2} = - 8 x$ .

Sol. The given equation is $y^{2} = - 8 x$ . Here, the coefficient of x is negative. Hence, the parabola opens towards the left. On comparing this equation with $y^{2} = - 4 ax$ , we obtain

$- 4 a = - 8 \Rightarrow a = 2$

∴ Coordinates of the focus $= (- a, 0) = (- 2, 0)$

Since the given equation involves $y^{2}$ , the axis of the parabola is the x-axis. Equation of directrix, $x = a$ i.e., $x = 2$ Length of latus rectum $= 4 a = 8$

$x^{2} = - 16 y$ .

Sol. The given equation is $x^{2} = - 16 y$ . Here, the coefficient of y is negative. Hence, the parabola opens downwards. On comparing this equation with $x^{2} = - 4 ay$ , we obtain $- 4 a = - 16 \Rightarrow a = 4$ $∴$ Coordinates of the focus $= (0, - a) = (0, - 4)$ Since the given equation involves $x^{2}$ , the axis of the parabola is the y-axis. Equation of directrix, $y = a$ i.e., $y = 4$ Length of latus rectum $= 4 a = 16$

$y^{2} = 10 x$

Sol. The given equation is $y^{2} = 10 x$ . Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with $y^{2} = 4 ax$ , we obtain $4 a = 10 \Rightarrow a = \frac{5}{2}$ ∴ Coordinates of the focus $= (a, 0) = (\frac{5}{2}, 0)$ Since the given equation involves $y^{2}$ , the axis of the parabola is the x-axis. Equation of directrix, $x = - a$ , i.e., $x = - \frac{5}{2}$ Length of latus rectum $= 4 a = 10$

$x^{2} = - 9 y$ .

Sol. The given equation is $x^{2} = - 9 y$ . Here, the coefficient of y is negative. Hence, the parabola opens downwards. On comparing this equation with $x^{2} = - 4 ay$ , we obtain $- 4 a = - 9 \Rightarrow a = \frac{9}{4}$ ∴ Coordinates of the focus $= (0, - a) = (0, - \frac{9}{4})$ Since the given equation involves $x^{2}$ , the axis of the parabola is the y-axis. Equation of directrix, $y = a$ , i.e., $y = \frac{9}{4}$ Length of latus rectum $= 4 a = 9$

In each of the Q. 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus $(6, 0)$ ; directrix $x = - 6$

Sol. Focus (6, 0); directrix, $x = - 6$ Since the focus lies on the $x$ -axis, the $x$ -axis is the axis of the parabola. Therefore, the equation of the parabola is either of the form $y^{2} = 4 ax$ or $y^{2} = - 4 ax$ . It is also seen that the directrix, $x = - 6$ is to the left of the y-axis, while the focus $(6, 0)$ is to the right of the y-axis. Hence, the parabola is of the form $y^{2} = 4 ax$ . Here, a = 6 Thus, the equation of the parabola is $y^{2} = 24 x$ .

8. Focus (0, -3); directrix $y = 3$

Sol. Focus = (0, -3); directrix y = 3 Since the focus lies on the y-axis, the y-axis is the axis of the parabola. Therefore, the equation of the parabola is either of the form $x^{2} = 4 ay$ or $x^{2} = - 4 ay$ . It is also seen that the directrix, $y = 3$ is above the x-axis, while the focus (0, -3) is below the x-axis. Hence, the parabola is of the form $x^{2} = - 4 ay$ . Here, a = 3 Thus, the equation of the parabola is $x^{2} = - 12 y$ .

9. Vertex $(0, 0)$ ; focus $(3, 0)$

Sol. Vertex (0, 0); focus $(3, 0)$ Since the vertex of the parabola is $(0, 0)$ and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form $y^{2} = 4 ax$ . Since the focus is $(3, 0), a = 3$ . Thus, the equation of the parabola is $y^{2} = 4 \times 3 \times x$ , i.e., $y^{2} = 12 x$

10. Vertex $(0, 0)$ focus $(- 2, 0)$

Sol. Vertex $(0, 0)$ focus $(- 2, 0)$ Since the vertex of the parabola is $(0, 0)$ and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form $y^{2} = - 4 ax$ . Since the focus is $(- 2, 0), a = 2$ . Thus, the equation of the parabola is $y^{2} = - 4 (2) x$ , i.e., $y^{2} = - 8 x$

Vertex $(0, 0)$ passing through $(2, 3)$ and axis is along x-axis Sol. Since the vertex is $(0, 0)$ and the axis of the parabola is the x-axis, the equation of the parabola is either of the form $y^{2} = 4 ax$ or $y^{2} = - 4 ax$ . The parabola passes through point (2, 3), which lies in the first quadrant. Therefore, the equation of the parabola is of the form $y^{2} = 4 ax$ , while point $(2, 3)$ must satisfy the equation $y^{2} = 4 ax$ . $∴ 3^{2} = 4 a (2) \Rightarrow a = \frac{9}{8}$

Thus, the equation of the parabola is

$y^{2} = 4 (\frac{9}{8}) x$

$\Rightarrow y^{2} = \frac{9}{2} x$ $\Rightarrow 2 y^{2} = 9 x$

12. Vertex $(0, 0)$ , passing through $(5, 2)$ and symmetric with respect to y-axis Sol. Since the vertex is $(0, 0)$ and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form $x^{2} = 4 ay$ or $x^{2} = - 4 ay$ . The parabola passes through point (5, 2), which lies in the first quadrant. Therefore, the equation of the parabola is of the form $x^{2} = 4 a y$ , while point $(5, 2)$ must satisfy the equation $x^{2} = 4 ay$ . $∴ (5)^{2} = 4 \times a \times 2$ $\Rightarrow 25 = 8 a$ $\Rightarrow a = \frac{25}{8}$

Thus, the equation of the parabola is

$x^{2} = 4 (\frac{25}{8}) y$

$\Rightarrow 2 x^{2} = 25 y$

EXERCISE - 10.3

In each of the Q. 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

$\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ Sol. The given equation is $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . Here, the denominator of $\frac{x ^{2}}{36}$ is greater than the denominator of $\frac{y ^{2}}{16}$ . Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. On comparing the given equation with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ $∴ c = a^{2} - b^{2} = 36 - 16 = 20 = 25$ we obtain $a = 6$ and $b = 4$ . Major axis is along the x-axis. Therefore, the coordinates of the foci are $(25, 0)$ and $(- 25, 0)$ . The coordinates of the vertices are $(6, 0)$ and (-6, 0). Length of major axis $= 2 a = 12$ Length of minor axis $= 2 b = 8$ Eccentricity, $e = \frac{c}{a} = \frac{2 5}{6} = \frac{5}{3}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 16}{6} = \frac{16}{3}$

$\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1$ . Sol. The given equation is $\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1$ Major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ we obtain $b = 2$ and $a = 5$ .

$∴ c = a^{2} - b^{2} = 25 - 4 = 21$

Therefore, tThe coordinates of the foci are $(0, 21)$ and $(0, - 21)$ . The coordinates of the vertices are $(0, 5)$ and (0,-5) Length of major axis $= 2 a = 10$ Length of minor axis $= 2 b = 4$ Eccentricity, $e = \frac{c}{a} = \frac{21}{5}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 4}{5} = \frac{8}{5}$

$\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ Sol. The given equation is $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ Here, the denominator of $\frac{x ^{2}}{16}$ is greater than the denominator of $\frac{y ^{2}}{9}$ . Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. On comparing the given equation with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ we obtain $a = 4$ and $b = 3$ . $∴ c = a^{2} - b^{2} = 16 - 9 = 7$

Therefore, The coordinates of the foci are $(\pm 7, 0)$ The coordinates of the vertices are (±4, 0) Length of major axis $= 2 a = 8$ Length of minor axis $= 2 b = 6$ Eccentricity, $e = \frac{c}{a} = \frac{7}{4}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2}$

$\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ . Sol. The given equation is $\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ Here, the denominator of $\frac{y ^{2}}{100}$ is greater than the denominator of $\frac{x ^{2}}{25}$ . Therefore, the major axis is along the y -axis, while the minor axis is along the x-axis. On comparing the given equation with $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ $∴ c = a^{2} - b^{2} = 100 - 25 = 75 = 53$ we obtain $b = 5$ and $a = 10$ . Therefore, The coordinates of the foci are $(0, \pm 53)$ . The coordinates of the vertices are (0, ±10). Length of major axis $= 2 a = 20$ Length of minor axis $= 2 b = 10$ Eccentricity, $e = \frac{c}{a} = \frac{5 3}{10} = \frac{3}{2}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 25}{10} = 5$
$\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1$ Sol. The given equation is $\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1$ Here, the denominator of $\frac{x ^{2}}{49}$ is greater than the denominator of $\frac{y ^{2}}{36}$ . Therefore, the major axis is along the x -axis, while the minor axis is along the y-axis. On comparing the given equation with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ we obtain $a = 7$ and $b = 6$ . $∴ c = a^{2} - b^{2} = 49 - 36 = 13$

Therefore, The coordinates of the foci are $(\pm 13, 0)$ . The coordinates of the vertices are (±7, 0). Length of major axis $= 2 a = 14$ Length of minor axis $= 2 b = 12$ Eccentricity, $e = \frac{c}{a} = \frac{13}{7}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 36}{7} = \frac{72}{7}$

6. $\frac{x ^{2}}{100} + \frac{y ^{2}}{400} = 1$

Sol. The given equation is $\frac{x ^{2}}{100} + \frac{y ^{2}}{400} = 1$ Here, the denominator of $\frac{y ^{2}}{400}$ is greater than the denominator of $\frac{x ^{2}}{100}$ . Therefore, the major axis is along the y -axis, while the minor axis is along the x-axis. On comparing the given equation with

$\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$

$∴ c = a^{2} - b^{2}$

$= 400 - 100 = 300 = 103$

we obtain $b = 10$ and $a = 20$ . Therefore, The coordinates of the foci are $(0, \pm 103)$ . The coordinates of the vertices are $(0, \pm 20)$ Length of major axis $= 2 a = 40$ Length of minor axis $= 2 b = 20$ Eccentricity, $e = \frac{c}{a} = \frac{10 3}{20} = \frac{3}{2}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 100}{20} = 10$

7. $36 x^{2} + 4 y^{2} = 144$

Sol. The given equation is $36 x^{2} + 4 y^{2} = 144$ . It can be written as

$36 x^{2} + 4 y^{2} = 144.$

or $\frac{x ^{2}}{4} + \frac{y ^{2}}{36} = 1$ or $\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{6 ^{2}} = 1$ Here, the denominator of $\frac{y ^{2}}{6 ^{2}}$ is greater than the denominator of $\frac{x ^{2}}{2 ^{2}}$ . Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing the given equation with

$\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$

$∴ c = a^{2} - b^{2} = 36 - 4 = 32 = 42$ we obtain $b = 2$ and $a = 6$ . Therefore, tThe coordinates of the foci are

$(0, \pm c) = (0, \pm 42) .$

The coordinates of the vertices are ( $0, \pm a$ )

$= (0, \pm 6) .$

Length of major axis $= 2 a = 12$ Length of minor axis $= 2 b = 4$ Eccentricity, $e = \frac{c}{a} = \frac{4 2}{6} = \frac{2 2}{3}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 4}{6} = \frac{4}{3}$

8. $16 x^{2} + y^{2} = 16$

Sol. The given equation is $16 x^{2} + y^{2} = 16$ . It can be written as

$16 x^{2} + y^{2} = 16$

or $\frac{x ^{2}}{1} + \frac{y ^{2}}{16} = 1$ or $\frac{x ^{2}}{1 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1$ Here, the denominator of $\frac{y ^{2}}{4 ^{2}}$ is greater than the denominator of $\frac{x ^{2}}{1}$ .

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing equation (1) with $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$

$∴ c = a^{2} - b^{2} = 16 - 1 = 15$ we obtain $b = 1$ and $a = 4$ . Therefore, the coordinates of the foci are $(0, \pm c) = (0, \pm 15)$ . The coordinates of the vertices are $(0, \pm a) = (0, \pm 4)$ . Length of major axis $= 2 a = 8$ Length of minor axis $= 2 b = 2$ Eccentricity, $e = \frac{c}{a} = \frac{15}{4}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 1}{4} = \frac{1}{2}$

9. $4 x^{2} + 9 y^{2} = 36$ . Sol. The given equation is $4 x^{2} + 9 y^{2} = 36$ . It can be written as $4 x^{2} + 9 y^{2} = 36$ or $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$

Here, the denominator of $\frac{x ^{2}}{3 ^{2}}$ is greater than the denominator of $\frac{y ^{2}}{2 ^{2}}$ . Therefore, the major axis is along the x-axis, while the minor axis is along the y -axis. On comparing the given equation with $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ we obtain $a = 3$ and $b = 2$ .

$∴ c = a^{2} - b^{2} = 9 - 4 = 5$ Therefore, the coordinates of the foci are $(\pm c, 0) = (\pm 5, 0)$ . The coordinates of the vertices are $(\pm a, 0) = (\pm 3, 0)$ . Length of major axis $= 2 a = 6$ Length of minor axis $= 2 b = 4$ Eccentricity, $e = \frac{c}{a} = \frac{5}{3}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$

In each of the following Q. 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Vertices $(\pm 5, 0)$ , foci $(\pm 4, 0)$ Sol. Vertices (±5, 0), foci (±4, 0) Here, the vertices are on the x-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $a = 5$ and $c = 4$ . It is known that $a^{2} = b^{2} + c^{2}$ $\Rightarrow c = a^{2} - b^{2} \Rightarrow c^{2} = a^{2} - b^{2}$ $\Rightarrow b^{2} = a^{2} - c^{2} \Rightarrow b = a^{2} - c^{2}$ $\Rightarrow b = 25 - 16 \Rightarrow b = 9 = 3$ Thus, the equation of the ellipse is $\frac{x ^{2}}{5 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1$ or $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$
Vertices $(0, \pm 13)$ , foci $(0, \pm 5)$ Sol. Vertices (0, ±13), foci (0, ±5) Here, the vertices are on the y-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $a = 13$ and $c = 5$ . It is known that $a^{2} = b^{2} + c^{2}$ $\Rightarrow c = a^{2} - b^{2} \Rightarrow c^{2} = a^{2} - b^{2}$ $\Rightarrow b = a^{2} - c^{2} \Rightarrow b = 169 - 25$ $\Rightarrow b = 144 = 12$

Thus, the equation of the ellipse is $\frac{x ^{2}}{1 2 ^{2}} + \frac{y ^{2}}{1 3 ^{2}} = 1$ or $\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$

Vertices (±6, 0), foci (±4, 0) Sol. Vertices (±6, 0), foci (±4, 0) Here, the vertices are on the x-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ where a is the semi-major axis.

Accordingly, $a = 6, c = 4$ . It is known that $a^{2} = b^{2} + c^{2}$

$\Rightarrow \Rightarrow \Rightarrow c = a^{2} - b^{2} b = a^{2} - c^{2} b = 20 \Rightarrow \Rightarrow c^{2} = a^{2} - b^{2} b = 36 - 16$

Thus, the equation of the ellipse is $\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{( 20 ) ^{2}} = 1$ or $\frac{x ^{2}}{36} + \frac{y ^{2}}{20} = 1$

13. Ends of major axis $(\pm 3, 0)$ , ends of minor axis $(0, \pm 2)$ . Sol. Ends of major axis (±3, 0), ends of minor axis $(0, \pm 2)$ Here, the major axis is along the x-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $a = 3$ and $b = 2$ . Thus, the equation of the ellipse is

$\frac{x ^{2}}{3 ^{2}} + \frac{y ^{2}}{2 ^{2}} = 1 i.e. \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$

Ends of major axis $(0, \pm 5)$ , ends of minor axis $(\pm 1, 0)$ Sol. Ends of major axis $(0, \pm 5)$ , ends of minor axis (±1, 0). Here, the major axis is along the y-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $a = 5$ and $b = 1$ . Thus, the equation of the ellipse is $\frac{x ^{2}}{1 ^{2}} + \frac{y ^{2}}{( 5 ) ^{2}} = 1$ or $\frac{x ^{2}}{1 ^{2}} + \frac{y ^{2}}{5} = 1$ .
Length of major axis 26, foci $(\pm 5, 0)$

Sol. Length of major axis = 26; foci = (±5, 0). Since the foci are on the x-axis, the major axis is along the x-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $2 a = 26$ $\Rightarrow a = 13$ and $c = 5$ . It is known that $a^{2} = b^{2} + c^{2}$ $\Rightarrow c = a^{2} - b^{2} \Rightarrow c^{2} = a^{2} - b^{2}$ $\Rightarrow b = a^{2} - c^{2} \Rightarrow b = 169 - 25$ $\Rightarrow b = 144 = 12$ Thus, the equation of the ellipse is

$\frac{x ^{2}}{1 3 ^{2}} + \frac{y ^{2}}{1 2 ^{2}} = 1 or \frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1$

Length of minor axis 16, foci $(0, \pm 6)$

Sol. Length of minor axis = 16; foci = (0, ±6). Since the foci are on the y -axis, the major axis is along the y-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $2 b = 16 \Rightarrow b = 8$ and $c = 6$ . It is known that $a^{2} = b^{2} + c^{2}$ $\Rightarrow c = a^{2} - b^{2} \Rightarrow a = b^{2} + c^{2}$ $\Rightarrow a = 64 + 36 \Rightarrow a = 100 = 10$ Thus, the equation of the ellipse is

$\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{1 0 ^{2}} = 1 or \frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1$

Foci $(\pm 3, 0), a = 4$

Sol. Foci $(\pm 3, 0), a = 4$ Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $c = 3$ and $a = 4$ . It is known that $a^{2} = b^{2} + c^{2}$

$\Rightarrow \Rightarrow \Rightarrow c = a^{2} - b^{2} 4^{2} = b^{2} + 3^{2} b^{2} = 16 - 9 = 7 \Rightarrow \Rightarrow b = a^{2} - c^{2} 16 = b^{2} + 9$

Thus, the equation of the ellipse is $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$

$b = 3, c = 4$ , centre at the origin; foci on the x axis. Sol. It is given that $b = 3, c = 4$ , centre at the origin; foci on the x axis. Since the foci are on the x-axis, the major axis is along the x-axis. Therefore, the equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semimajor axis. Accordingly, $b = 3, c = 4$ . It is known that $a^{2} = b^{2} + c^{2}$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow c = a^{2} - b^{2} a^{2} = b^{2} + c^{2} a = 25 a = 9 + 16 \Rightarrow \Rightarrow \Rightarrow$

Thus, the equation of the ellipse is

$\frac{x ^{2}}{5 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1 or \frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$

Centre at (0, 0), major axis on the y-axis and passes through the points $(3, 2)$ and (1, 6). Sol. Since the centre is at $(0, 0)$ and the major axis is on the y-axis, the equation of the ellipse will be of the form

$\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{a ^{2}} = 1$

Where, a is the semi-major axis The ellipse passes through points $(3, 2)$ and (1, 6). Hence,

$\frac{9}{b ^{2}} + \frac{4}{a ^{2}} = 1$

$\frac{1}{b ^{2}} + \frac{36}{a ^{2}} = 1$

On solving equations (2) and (3), we obtain $b^{2} = 10$ and $a^{2} = 40$ . Thus, the equation of the ellipse is

$\frac{x ^{2}}{10} + \frac{y ^{2}}{40} = 1$

or $4 x^{2} + y^{2} = 40$

Major axis on the x -axis and passes through the points $(4, 3)$ and (6, 2). Sol. Since the major axis is on the x-axis, the equation of the ellipse will be of the form

$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Where, a is the semi-major axis The ellipse passes through points $(4, 3)$ and (6, 2). Hence,

$\frac{16}{a ^{2}} + \frac{9}{b ^{2}} = 1 \frac{36}{a ^{2}} + \frac{4}{b ^{2}} = 1$

On solving equations (2) and (3), we obtain $a^{2} = 52$ and $b^{2} = 13$ . Thus, the equation of the ellipse is

$\frac{x ^{2}}{52} + \frac{y ^{2}}{13} = 1 or x^{2} + 4 y^{2} = 52$

EXERCISE - 10.4

In each of the Q. 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

$\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ Sol. The given equation is

$\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1 or \frac{x ^{2}}{4 ^{2}} - \frac{y ^{2}}{3 ^{2}} = 1.$

On comparing this equation with the standard equation of hyperbola i.e., $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , we obtain $a = 4$ and $b = 3$ . We know that $a^{2} + b^{2} = c^{2}$

$\Rightarrow c = a^{2} + b^{2} \Rightarrow c^{2} = 4^{2} + 3^{2}$ $\Rightarrow c = 16 + 9 \Rightarrow a = 25$ $\Rightarrow a = 5$ Therefore, The coordinates of the foci are (±5, 0). The coordinates of the vertices are (±4, 0). Eccentricity, $e = \frac{c}{a} = \frac{5}{4}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2}$

$\frac{y ^{2}}{9} - \frac{x ^{2}}{27} = 1$ Sol. The given equation

$\frac{y ^{2}}{9} - \frac{x ^{2}}{27} = 1 or \frac{y ^{2}}{3 ^{2}} - \frac{x ^{2}}{( 27 ) ^{2}} = 1$

On comparing this equation with the standard equation of hyperbola $\frac{y ^{2}}{a ^{2}} + \frac{x ^{2}}{b ^{2}} = 1$ , i.e., we obtain $a = 3$ and $b = 27$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow c = a^{2} + b^{2}$ $∴ c = 3^{2} + (27)^{2} = 9 + 27 = 36$ $\Rightarrow c = 6$

Therefore, The coordinates of the foci are (0, ±6). The coordinates of the vertices are (0, ±3). Eccentricity, $e = \frac{c}{a} = \frac{6}{3} = 2$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 27}{3} = 18$

$9 y^{2} - 4 x^{2} = 36$ Sol. The given equation is $9 y^{2} - 4 x^{2} = 36$ . It can be written as $9 y^{2} - 4 x^{2} = 36$ or $\frac{y ^{2}}{4} - \frac{x ^{2}}{9} = 1$ or $\frac{y ^{2}}{2 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1$ On comparing equation (1) with the standard equation of hyperbola $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ i.e.,, we obtain $a = 2$ and $b = 3$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow c = a^{2} + b^{2} \Rightarrow c = 4 + 9$ $\Rightarrow c = 13$ Therefore, the coordinates of the foci are $(0, \pm 13)$ The coordinates of the vertices are $(0, \pm 2)$ Eccentricity, $e = \frac{c}{a} = \frac{13}{2}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 9}{2} = 9$
$16 x^{2} - 9 y^{2} = 576$ Sol. The given equation is $16 x^{2} - 9 y^{2} = 576$ . It can be written as

$16 x^{2} - 9 y^{2} = 576$

$\Rightarrow \frac{x ^{2}}{36} - \frac{y ^{2}}{64} = 1 \Rightarrow \frac{x ^{2}}{6 ^{2}} - \frac{y ^{2}}{8 ^{2}} = 1$ On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , we obtain $a = 6$ and $b = 8$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow c = a^{2} + b^{2} \Rightarrow c = 36 + 64$ $\Rightarrow c = 100 \Rightarrow c = 10$ Therefore, the coordinates of the foci are (±10, 0). The coordinates of the vertices are (±6, 0). Eccentricity, $e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 64}{6} = \frac{64}{3}$

$5 y^{2} - 9 x^{2} = 36$

Sol. The given equation is $5 y^{2} - 9 x^{2} = 36$

$\Rightarrow \frac{y ^{2}}{( \frac{36}{5} )} - \frac{x ^{2}}{4} = 1 \Rightarrow \frac{y ^{2}}{( \frac{6}{5} ) ^{2}} - \frac{x ^{2}}{2 ^{2}} = 1$

On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ , we obtain $a = \frac{6}{5}$ and $b = 2$ . We know that $a^{2} + b^{2} = c^{2}$ .

$\Rightarrow ∴ \Rightarrow c = a^{2} + b^{2} c = \frac{36}{5} + 4 = \frac{56}{5} c = \frac{2 14}{5}$

Therefore, the coordinates of the foci are $(0, \pm \frac{2 14}{5})$ The coordinates of the vertices are $(0, \pm \frac{6}{5})$ Eccentricity, $e = \frac{c}{a} = \frac{( \frac{2 14}{5} )}{( \frac{6}{5} )} = \frac{14}{3}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 4}{( \frac{6}{5} )} = \frac{4 5}{3}$

6. $49 y^{2} - 16 x^{2} = 784$

Sol. The given equation is $49 y^{2} - 16 x^{2} = 784$ It can be written as $49 y^{2} - 16 x^{2} = 784$ or $\frac{y ^{2}}{16} - \frac{x ^{2}}{49} = 1$ or $\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{7 ^{2}} = 1$ On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ , we obtain $a = 4$ and $b = 7$ . We know that $a^{2} + b^{2} = c^{2}$ .

$\Rightarrow c = a^{2} + b^{2} \Rightarrow c = 16 + 49 \Rightarrow c = 65$

Therefore, The coordinates of the foci are $(0, \pm 65)$ . The coordinates of the vertices are (0, ±4). Eccentricity, $e = \frac{c}{a} = \frac{65}{4}$ Length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{2 \times 49}{4} = \frac{49}{2}$

In each of the Q. 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Vertices (±2, 0), foci (±3, 0)

Sol. Vertices (±2, 0), foci (±3, 0) Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Since the vertices are $(\pm 2, 0), a = 2$ . Since the foci are (±3, 0), $c = 3$ . We know that $a^{2} + b^{2} = c^{2}$ .

$∴ 2^{2} + b^{2} = 3^{2} b^{2} = 9 - 4 = 5$

Thus, the equation of the hyperbola is $\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1$

Vertices (0, ±5), foci $(0, \pm 8)$

Sol. Vertices (0, ±5), foci (0, ±8) Here, the vertices are on the y-axis. Therefore, the equation of the hyperbola is of the form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Since the vertices are $(0, \pm 5), a = 5$ . Since the foci are (0, ±8), $c = 8$ .

We know that $a^{2} + b^{2} = c^{2}$ .

$∴ 5^{2} + b^{2} = 8^{2}$

$b^{2} = 64 - 25 = 39$

Thus, the equation of the hyperbola is

$\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1$

Vertices $(0, \pm 3)$ , foci $(0, \pm 5)$ Sol. Vertices (0, ±3), foci (0, ±5) Here, the vertices are on the y-axis. Therefore, the equation of the hyperbola is of the form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Since the vertices are $(0, \pm 3), a = 3$ . Since the foci are (0, ±5), $c = 5$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow b = c^{2} - a^{2}$ $\Rightarrow b = 25 - 9 \Rightarrow b = 16 = 4$ Thus, the equation of the hyperbola is $\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1$
Foci $(\pm 5, 0)$ , the transverse axis is of length 8. Sol. Foci (±5, 0), the transverse axis is of length 8. Here, the foci are on the x-axis. Therefore, the equation of the hyperbola is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Since the foci are (±5,0), $c = 5$ . Since the length of the transverse axis is 8, $2 a = 8 \Rightarrow a = 4$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow b = c^{2} - a^{2} \Rightarrow b = 25 - 16$ $\Rightarrow b = 9 = 3$

Thus, the equation of the hyperbola is $\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1$ .

Foci $(0, \pm 13)$ , the conjugate axis is of length 24. Sol. Foci $(0, \pm 13)$ , the conjugate axis is of length 24. Here, the foci are on the y -axis. Therefore, the equation of the hyperbola is of the form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Since the foci are (0, ±13), $c = 13$ . Since the length of the conjugate axis is $24, 2 b = 24$ $\Rightarrow b = 12$ . We know that $a^{2} + b^{2} = c^{2}$ . $\Rightarrow a = c^{2} - b^{2}$ $\Rightarrow a = 169 - 144 \Rightarrow a = 25 = 5$ Thus, the equation of the hyperbola is $\frac{y ^{2}}{25} - \frac{x ^{2}}{144} = 1$
Foci $(\pm 35, 0)$ , the latus rectum is of length 8. Sol. Foci $(\pm 35, 0)$ , the latus rectum is of length 8. Here, the foci are on the x-axis. Therefore, the equation of the hyperbola is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Since the foci are $(\pm 35, 0), c = \pm 35$ Length of latus rectum $= 8$ $\Rightarrow \frac{2 b ^{2}}{a} = 8 \Rightarrow b^{2} = 4 a$ We know that $a^{2} + b^{2} = c^{2}$ . $∴ a^{2} + 4 a = 45$ $\Rightarrow a^{2} + 4 a - 45 = 0$ $\Rightarrow a^{2} + 9 a - 5 a - 45 = 0$ $\Rightarrow (a + 9) (a - 5) = 0$ $\Rightarrow a = - 9, 5$ Since a is non-negative, $a = 5$ . $∴ b^{2} = 4 a = 4 \times 5 = 20$ Thus, the equation of the hyperbola is $\frac{x ^{2}}{25} - \frac{y ^{2}}{20} = 1$ .
Foci $(\pm 4, 0)$ , the latus rectum is of length 12 Sol. Foci (±4, 0), the latus rectum is of length 12. Here, the foci are on the x-axis. Therefore, the equation of the hyperbola is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Since the foci are (±4, 0), $c = 4$ . Length of latus rectum = 12 $\Rightarrow \frac{2 b ^{2}}{a} = 12$ $\Rightarrow b^{2} = 6 a$

We know that $a^{2} + b^{2} = c^{2}$ .

$∴ a^{2} + 6 a = 16$ $\Rightarrow a^{2} + 6 a - 16 = 0$ $\Rightarrow a^{2} + 8 a - 2 a - 16 = 0$ $\Rightarrow (a + 8) (a - 2) = 0$ $\Rightarrow a = - 8, 2$

Since a is non-negative, $a = 2$ .

$∴ b^{2} = 6 a = 6 \times 2 = 12$

Thus, the equation of the hyperbola is $\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1$

Vertices $(\pm 7, 0), e = \frac{4}{3}$ Sol. Vertices $(\pm 7, 0), e = \frac{4}{3}$

Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Since the vertices are $(\pm 7, 0), a = 7$ . It is given that $e = \frac{4}{3}$

$∴ \frac{c}{a} = \frac{4}{3}$ $[e = \frac{c}{a}]$ $\Rightarrow \frac{c}{7} = \frac{4}{3}$ $\Rightarrow c = \frac{28}{3}$

We know that $a^{2} + b^{2} = c^{2}$ .

$∴ 7^{2} + b^{2} = (\frac{28}{3})^{2}$ $\Rightarrow b^{2} = \frac{784}{9} - 49$ $\Rightarrow b^{2} = \frac{784 - 441}{9} = \frac{343}{9}$ Thus, the equation of the hyperbola is $\frac{x ^{2}}{49} - \frac{9 y ^{2}}{343} = 1$

Foci $(0, \pm 10)$ , passing through $(2, 3)$ . Sol. Foci $(0, \pm 10)$ , passing through $(2, 3)$ Here, the foci are on the y -axis. Therefore, the equation of the hyperbola is of the form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Since the foci are $(0, \pm 10), c = 10$ We know that $a^{2} + b^{2} = c^{2}$ . $∴ a^{2} + b^{2} = 10$ $\Rightarrow b^{2} = 10 - a^{2}$ .....(1) Since the hyperbola passes through point $(2, 3), \frac{9}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ .....(2) From equations (1) and (2), we obtain $\frac{9}{a ^{2}} - \frac{4}{( 10 - a ^{2} )} = 1$ $9 (10 - a^{2}) - 4 a^{2} = a^{2} (10 - a^{2})$ $\Rightarrow 90 - 9 a^{2} - 4 a^{2} = 10 a^{2} - a^{4}$ $\Rightarrow a^{4} - 23 a^{2} + 90 = 0$ $\Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0$ $\Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0$ $\Rightarrow (a^{2} - 18) (a^{2} - 5) = 0$ $\Rightarrow a^{2} = 18$ or 5 In hyperbola, $c > a$ , i.e., $c^{2} > a^{2}$ ⇒ $(10)^{2} > 5$ $\Rightarrow 10 > 5$ $∴ a^{2} = 5$ $\Rightarrow b^{2} = 10 - a^{2} = 10 - 5 = 5$ Thus, the equation of the hyperbola is $\frac{y ^{2}}{5} - \frac{x ^{2}}{5} = 1$

MISCELLANEOUS EXERCISE

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Sol. This can be diagrammatically represented as

Class-11-maths-chapter-10-mis-exer-ques-1

The equation of the parabola is of the form $y^{2} = 4 ax$ (as it is opening to the right). Since the parabola passes through point $A (5, 10), 1 0^{2} = 4 a (5)$ ⇒

$100 = 20 a \Rightarrow a = \frac{100}{20} = 5$

Therefore, the focus of the parabola is $(a, 0) = (5, 0)$ , which is the mid-point of the diameter. Hence, the focus of the reflector is at the midpoint of the diameter.

An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola? Sol. This can be diagrammatically represented as

ques-2-mis-exer-class-11-maths-chapter-10

The equation of the parabola is of the form $x^{2} = 4 ay$ (as it is opening upwards). It can be clearly seen that the parabola passes through point $(\frac{5}{2}, 10)$ .

$\Rightarrow (\frac{5}{2})^{2} = 4 a (10) a = \frac{25}{4 \times 4 \times 10} = \frac{5}{32}$

Therefore, the arch is in the form of a parabola whose equation is $x^{2} = \frac{5}{8} y$ . When $y = 2 m, x^{2} = \frac{5}{8} \times 2$ $\Rightarrow x^{2} = \frac{5}{4} \Rightarrow x = \frac{5}{4} m$ $∴ AB = 2 \times \frac{5}{4} m = 2 \times 1.118 m$ (approx.) $= 2.23 m$ (approx.) Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Sol. This can be diagrammatically represented as

ques-3-mis-exer-chapter-10-maths-class-11

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable. DF is the supporting wire attached to the roadway, 18 m from the middle. Here, $AB = 30 m, OC = 6 m$ and $BC = \frac{100}{2} = 50 m$ . The equation of the parabola is of the form $x^{2}$ =4ay (as it is opening upwards). The coordinates of point A are $(50, 30 - 6) =$ $(50, 24)$ . Since A $(50, 24)$ is a point on the parabola,

$(50)^{2} = 4 a (24) \Rightarrow a = \frac{50 \times 50}{4 \times 24} = \frac{625}{24}$

$∴$ Equation of the parabola

$x^{2} = 4 \times \frac{625}{24} \times y or 6 x^{2} = 625 y$

The x -coordinate of point D is 18 . Hence, at $x = 18$ ,

$6 (18)^{2} = 625 y$

$\Rightarrow y = \frac{6 \times 18 \times 18}{625} = 3.11 (approx.)$

$∴ DE = 3.11 m$

$DF = DE + EF = 3.11 m + 6 m = 9.11 m$

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end. Sol. The semi-ellipse can be diagrammatically represented as

class-10-maths-chapter-5-mis-exer-ques-4

The equation of the semi-ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, y \geq 0$ where a is the semimajor axis Accordingly, $2 a = 8 \Rightarrow a = 4, b = 2$ Therefore, the equation of the semi-ellipse is

$\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1, y \geq 0$

Let A be a point on the major axis such that $AB = 1.5 m$ . Draw AC ⟂ OB. $OA = (4 - 1.5) m = 2.5 m$ The x-coordinate of point C is 2.5. On substituting the value of x with 2.5 in equation (1), we obtain

$\frac{( 2.5 ) ^{2}}{16} + \frac{y ^{2}}{4} = 1 \Rightarrow \frac{6.25}{16} + \frac{y ^{2}}{4} = 1$

$\Rightarrow y^{2} = 4 (1 - \frac{6.25}{16}) \Rightarrow y^{2} = 4 (\frac{9.75}{16})$ $\Rightarrow y^{2} = 2.4375 \Rightarrow y = 1.56$ (approx.) $∴ AC = 1.56 m$ Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis. Sol. Let AB be the rod making an angle $θ$ with OX and $P (x, y)$ be the point on it such that $AP = 3 cm$ . Then, $PB = AB - AP = (12 - 3) cm = 9 cm$ $[AB = 12 cm]$ From P , draw $PQ ⊥ OY$ and $PR ⊥ OX$ .

mis-exer-ques-5-maths-class-11-chapter-10

$In △ PBQ, cos θ = \frac{PQ}{PB} = \frac{x}{9} In △ PR A, sin θ = \frac{PR}{P A} = \frac{y}{3}$

Since, $sin^{2} θ + cos^{2} θ = 1$

$(\frac{y}{3})^{2} + (\frac{x}{9})^{2} = 1 or \frac{x ^{2}}{81} + \frac{y ^{2}}{9} = 1$

Thus, the equation of the locus of point P on the rod is $\frac{x ^{2}}{81} + \frac{y ^{2}}{9} = 1$

Find the area of the triangle formed by the lines joining the vertex of the parabola $x^{2} =$ 12y to the ends of its latus rectum. Sol. The given parabola is $x^{2} = 12 y$ . On comparing this equation with $x^{2} = 4 ay$ , we obtain $4 a = 12 \Rightarrow a = 3$ $∴$ The coordinates of foci are $S (0, a) = S (0, 3)$ Let AB be the latus rectum of the given parabola. The given parabola can be roughly drawn as

class-11-maths-chapter-10-ques-6-mis-exer

At $y = 3, x^{2} = 12$ (3)

$\Rightarrow x^{2} = 36 \Rightarrow x = \pm 6$

∴ The coordinates of A are (-6, 3), while the coordinates of B are (6, 3). Therefore, the vertices of $△ OAB$ are $O (0, 0), A (- 6, 3)$ , and B (6, 3).

$AB = (6 + 6)^{2} + (3 - 3)^{2}$

$AB = (12)^{2} = 12 = \frac{1}{2} \times AB \times OS = \frac{1}{2} \times 12 \times 3 = 6 \times 3 = 18 unit^{2}$

Thus, the required area of the triangle is 18 unit $^{2}$ .

A man running a racecourse notes that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m . find the equation of the posts traced by the man.

Sol. Let A and B be the positions of the two flag posts and $P (x, y)$ be the position of the man. Accordingly, $PA + PB = 10$ . We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse. The ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , where a is the semi-major axis. Accordingly, $2 a = 10 \Rightarrow a = 5$ Distance between the foci $(2 c) = 8 \Rightarrow c = 4$ On using the relation $c = a^{2} - b^{2}$

$4 = 25 - b^{2} \Rightarrow b^{2} = 25 - 16 = 9 \Rightarrow 16 = 25 - b^{2} \Rightarrow b = 3$

Thus, the equation of the path traced by the man is $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ .

An equilateral triangle is inscribed in the parabola $y^{2} = 4 ax$ , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. Sol. Let OAB be the equilateral triangle inscribed in parabola $y^{2} = 4 ax$ . Let AB intersect the x-axis at point C.

Let $OC = k$

From the equation of the given parabola, we have

$y^{2} = 4 ak \Rightarrow y = \pm 2 ak$

∴ The respective coordinates of points A and B are

$(k, 2 ak) and (k, - 2 ak) AB = CA + CB = 2 ak + 2 ak = 4 ak$

Since OAB is an equilateral triangle,

$OA^{2} = AB^{2} .$

$∴ k^{2} + (2 ak)^{2} = (4 ak)^{2}$ $\Rightarrow k^{2} + 4 ak = 16 ak \Rightarrow k^{2} = 12 ak$ $\Rightarrow k = 12 a$ $∴ AB = 4 ak = 4 a \times 12 a$

$= 4 12 a^{2} = 83 a$

Thus, the side of the equilateral triangle inscribed in parabola $y^{2} = 4$ ax is $83 a$ .

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Explore Class 11 Maths NCERT Solutions chapter-wise, with solved exercises, important formulas, and easy explanations to help understand and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

4.0Class 11 Maths Chapter 10 Conic Sections : Exercise-Wise Questions and Topics

Exercise	Questions & Solutions	Important Concepts Covered
Exercise 10.1	15 Questions & Solutions	Circle, equation of a circle, centre and radius, and circle-related problems
Exercise 10.2	12 Questions & Solutions	Parabola, focus, axis, directrix, latus rectum, and equations of parabolas
Exercise 10.3	20 Questions & Solutions	Ellipse, foci, vertices, major and minor axes, eccentricity, latus rectum, and equations of ellipses
Exercise 10.4	15 Questions & Solutions	Hyperbola, foci, vertices, eccentricity, latus rectum, and equations of hyperbolas
Miscellaneous Exercise	8 Questions & Solutions	Mixed questions covering circles, parabolas, ellipses, and hyperbolas

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 9 (Straight Lines)

Clear Understanding of Coordinate Geometry: Learn how points, slopes, intercepts, and straight lines are represented on the Cartesian plane through clear explanations and graphical examples.
Different Forms of the Equation of a Line: Understand the slope-intercept, point-slope, two-point, intercept, and normal forms of a straight line and learn when to use each form.
Step-by-Step Slope Calculations: Learn how to find the slope of a line from two points and understand the relationship between slope and the angle of inclination, including cases involving obtuse angles.
Intercepts and Graphical Representation: Understand how to find the (x)-intercept and (y)-intercept of a line and represent the equation correctly on a coordinate plane.
Distance from a Point to a Line: Learn how to calculate the perpendicular distance between a point and a straight line using the appropriate formula and clear calculation steps.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with step-by-step solutions to help students understand and apply the concepts of Straight Lines.

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Frequently Asked Questions

Why are NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections important for students?

How do NCERT Solutions for Class 11 Maths Chapter 10 help in board and JEE preparation?

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Are NCERT Solutions for Class 11 Maths Chapter 10 by ALLEN useful for exam preparation?

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NCERT Solutions Class 11 Maths Chapter 10 – Conic Sections

1.0Class 11 Maths Chapter 10 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 9

EXERCISE - 10.1

EXERCISE - 10.2

EXERCISE - 10.3

EXERCISE - 10.4

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Class 11 Maths Chapter 10 Conic Sections : Exercise-Wise Questions and Topics

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 9 (Straight Lines)

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