Conic Sections

Conic Sections are the curves obtained by an interesting double-napped right circular cone by a plane. They include circles, ellipses, parabolas, and hyperbolas, each with unique properties. Circles have equidistant points from the center, ellipses possess two foci, parabolas feature a single focus and directrix, while hyperbolas have two separate branches.

1.0What are Conic Sections?

Conic sections are the curves produced when a right circular cone is intersected by a plane.

Geometrical Interpretation

Let l be a fixed vertical line and m be another line intersecting it at a fixed-Point V and inclined to it at an angle 𝛼. Now, If we rotate a line m around another line l while maintaining a constant angle between them, the surface generated is a double-napped right circular hollow cone. When a plane intersects this cone, the resulting curve is known as a conic section. Therefore, conic sections are the curves formed by the intersection of a plane with a right circular cone.

2.0Types of Conic Section

When a double-napped right circular cone is intersected by a plane (other than the vertex) then we have the following situations: (Here, 𝛽 represents the angle between the intersecting plane and the vertical axis of the cone.)

1. When 𝛽 = 90^o, then it is a circle. 
2. When 𝛼 ˂ 𝛽 ˂ 90^o, then it is an ellipse.
3. When 𝛽 = 𝛼, then it is a parabola.
4. When 0 ≤ 𝛽 ˂ 𝛼, then it is a hyperbola.

3.0Circles

A circle is the locus (set) of points (x, y), in the plane, that are all at a fixed distance, called the radius(r), from a fixed point, the center (u, v):

Equation of the Circle

The equation of a circle centered at any point in the coordinate system- the general equation- is obtained by expressing the radius r in terms of the distance formula; thus, 

leading to $r=\sqrt{(x-u{)}^2+(y-v{)}^2}$ ,

$(x-u{)}^2+(y-v{)}^2=r^2$

The equation of a circle centered at the origin of the coordinate system – the central equation – is 

$x^2+y^2=r^2$

4.0Ellipse

An ellipse is the locus(set) of all points in a plane for which the sum of their distance from two fixed points is constant.

Equation of an Ellipse 

The general equation of an ellipse with its center at the origin, semi-major axis ‘a’ along the x-axis and semi-minor axis ‘b’ along the y-axis is given by:

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

This equation represents all points on the ellipse where the sum of the squares of their distances from the center is equal to 1. The measure of the semi-major axis ‘a’ and semi-minor axis ‘b’ determine the size and shape of the ellipse.

Eccentricity of Ellipse

5.0Parabola

A Parabola is the locus of points (x, y), in the plane, whose distances to a given straight line (the directrix) and from a given point (focus) outside the line are equal.

The parabola is symmetrical with respect to an axis passing through the focus and perpendicular to the directrix. The vertex of the parabola is defined to be located at the midpoint of the segment of the axis extending from the focus to the directrix.

Equation of a Parabola

For the general equation of a parabola whose axis is parallel to or coincides with the y axis, consider the graph of the parabola below, whose vertex is at (u, v)

By definition of the parabola, we have

$\sqrt{(x-u{)}^2+[\mathrm{y}-(\mathrm{v}+\mathrm{p}){]}^2}=y-(v-p)$

$(x-u{)}^2=4p(y-v)$

For the central and general equations of a parabola whose axis coincides with or is parallel to the x-axis, one follows a similar pattern.

6.0Hyperbola

A Hyperbola is the locus(set) of Points, in the plane where the absolute difference in their  distances from two fixed points (foci) remain constant.

Equation of a Hyperbola

If, instead, the center of the hyperbola is at a point (u, v), we have the general equations:

$\frac{(x-u{)}^2}{a^2}-\frac{(y-v{)}^2}{b^2}=1$

foci on a line parallel to or coinciding with the x-axis;

$\frac{(y-u{)}^2}{a^2}-\frac{(x-v{)}^2}{b^2}=1$

foci on a line parallel to or coinciding with the y-axis.

For the general equation of a hyperbola with foci on a line coinciding with or parallel to the x-axis, consider the graph of the hyperbola below, centered at a point (u, v):

From the definition of the hyperbola, we have

$\left|\sqrt{[x-(u-c){]}^2+(y-v{)}^2}-\sqrt{[x-(u+c){]}^2+(y-v{)}^2}\right|=2a\mid$

which, after some algebra, may be rewritten

$\frac{(x-u{)}^2}{a^2}-\frac{(y-v{)}^2}{b^2}=1$

7.0Solved Example Based on Conic Section

Example 1: Find the equation of the circle whose center is (0,3) and radius is 3. 

Solution:

The equation of the circle is $(x-0{)}^2+(y-3{)}^2=3^2$

⇒ $x^2+y^2-6y=0$

Example 2: Find the equation of the circle which passes through (1, -1) and two of its diameters are x+2 y-5=0  and x-y+1=0 

Solution:

Let P be the point of intersection of the lines 

x+2 y-5=0 ...(i)   

and x-y+1=0 ...(ii)

Solving (i) and (ii), we get x=1, y=2 . So, the coordinates of the center are (1,2). since circle passes through (1, –1) 

so $\text{ radius }=\sqrt{(1-1{)}^2+(2+1{)}^2}\Rightarrow \text{ radius }=3$

Hence the equation of the required circle is$(x-1{)}^2+(y+2{)}^2=9$ 

Example 3: Identify the conic  x² + 2y² – 2xy + 8x + 10y – 36 = 0    

Solution:

a = 1, b = 1, h = –1, g = 5, f = 9 c = –46 

Now, Δ =  abc + 2fgh – af² – bg² – ch² ≠ 0 

and h² < ab, hence ellipse

Example 4: Find all the parameters of the following parabolas and draw their graphs. 

(i) y² = 16x

Solution:

a = 4, 

focus = (4, 0), 

vertex = (0, 0),

directrix :  x + 4 = 0

axis : y = 0,

Latus rectum equation x = 4

Length of Latus Rectum = 16,

Endpoints of latus rectum (4, 8) & (4, –8)           

Example 5: Find the equation of the parabola

(a) whose vertex is the point (4, –3) and whose latus rectum is 4 and whose axis is parallel to the x-axis and focus lies on the right side of the vertex.

Solution:

vertex is (4, – 3) and L₁L₂ = 4

∴ (y + 3)² = 4(x – 4)

y^2+6 y+9=4 x-16

y^2-4 x+6 y+25=0

(b) passing through the point (–4, –7) and whose directrix is parallel to x-axis and whose vertex is the point (4, –3).

Example 6: For what values of ‘a’ the point P(\alpha, \alpha) lies inside, on or outside the parabola

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

Solution:

Given equation can be written as y^2-4 y-4 x+16=0 

$PointP(\alpha ,\alpha )$ lies inside parabola if ${\alpha }^2-8\alpha +16<0$ 

⇒ $(\alpha -4{)}^2<0$ ⇒ no such α exists.

Point $P(\alpha ,\alpha )$ lies on parabola if $(\alpha -4{)}^2=0\Rightarrow \alpha =4$ 

Point P(α, α) lies outside parabola if $(\alpha -4{)}^2>0\Rightarrow \alpha \in R-\{4\}$

Example 7: Find the equation of an ellipse whose focus is (2, 3) eccentricity is 1/3 and the directrix x+2 y-1=0 

Solution:

Let P(x,y) be any point on the ellipse

whose focus is s(2, 3) and the directrix is x + 2y –1 = 0

PM perpendicular from P(x, y) on the directrix x + 2y –1 = 0 

Then by definition SP = $ePM\Rightarrow (SP{)}^2=e^2(PM{)}^2$

⇒ $(x-2{)}^2+(y-3{)}^2=\frac{1}{9}\left\{\frac{x+2y-1}{\sqrt{5}}\right\}$

⇒ $45\left(x^2+y^2-4x-6y+13\right)=x^2+4y^2+1+4xy-2x-4y$  

⇒ $44x^2+41y^2-4xy-178x-266y+584=0$

which is the required equation of the ellipse.

Example 8: The equation of the standard ellipse with respect to coordinate axes whose minor axis is equal to the distance between its foci and whose L R=10 , will be-

(A) $2x^2+y^2=100$    (B) $x^2+2y^2=100$

(C) $2x^2+3y^2=80$ (D) none of these

Ans. (A, B)

Solution:

When a > b 

As given  $2b=2ae\Rightarrow b=ae$ ...(i)

Also  $\frac{2b^2}{a}=10\Rightarrow b^2=5a$ ...(ii)

Now since $b^2=a^2-a^2{e}^2\Rightarrow b^2=a^2-b^2$ [From (i)]

⇒ 2b² =a² ...(iii) 

(ii), (iii) ⇒a² = 100, b² = 50    

Hence equation of the ellipse will be $\frac{x^2}{100}+\frac{y^2}{50}=1\Rightarrow x^2+2y^2=100$ 

Similarly, when a < b  then required ellipse is 2x² + y² = 100 

Example 9: Find the equation of a hyperbola whose focus is (1, 2) eccentricity is $\sqrt{3}$ and the directrix 2 x+3 y=0 .

Solution:

Let P(x,y) be any point on the hyperbola

Whose focus is S(1, 2) and the directrix is 2x + 3y = 0 

PM is perpendicular from P(x,y) on the directrix 2x + 3y = 0 

Then by definition SP = e.PM 

$(SP{)}^2=e^2(PM{)}^2\Rightarrow (x-1{)}^2+(y-2{)}^2=3{\left\{\frac{2x+3y}{\sqrt{13}}\right\}}^2$

⇒ $13\left(x^2+y^2-2x-4y+5\right)=3\left(4x^2+9y^2+12xy\right)$

⇒ $x^2-14y^2-36xy-26x-52y+65=0$ 

which is the required equation of the hyperbola.

Example 10: Find all parameters of hyperbola $3x^2-6y^2=-18$.

Solution:

$\frac{x^2}{6}-\frac{y^2}{3}=-1,$ Conjugate Hyperbola

$a^2=6,b^2=3$

Transverse Axis =$2b=2\sqrt{3}$

Conjugate Axis = $2\sqrt{6}$ 

$e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}$

Foci: $(0,\pm be)\equiv (0,\pm 3)$ 

Vertices: $(0,\pm b)\equiv (0,\pm \sqrt{3})$  

Latus Rectum = $\frac{2a^2}{b}=4\sqrt{3}$ 

Directrix: y= $\pm \frac{b}{e}\Rightarrow y=\pm 1$

8.0Practice Question Based on Conic Sections

1. The radius of the circle passing through given points (1,2),(5,2) \&(5,-2) is

(A)  $5\sqrt{2}$ (B) $2\sqrt{5}$ (C) $3\sqrt{2}$ (D) $2\sqrt{2}$

Ans: D

2. Find the equation to the circle which passes through the origin and cuts off intercepts equal to 3 & 4 from the positive axes.

(A) $x^2+y^2+3x+4y=0$ (B) $x^2+y^2-3x+4y=0$

(C) $x^2+y^2+3x-4y=0$ (D) $x^2+y^2-3x-4y=0$

Ans: D

3. Find the equation of the circle that passes through the points (1, 0), (–1, 0) and (0, 1) .

(A) $x^2+y^2=2$ (B) $x^2+y^2=1$

(C) $x^2-2x+y^2=0$ (D) $x^2+2x+y^2=2$

Ans: B

4. Find the length of latus rectum of parabola y² = 40x 

(A) 10 (B) 20 (C) 40 (D) 60 

Ans:  C

5. Find the equation of the parabola with its focus (–1, –2) and the directrix is x – 2y + 3 = 0 .

(A) $x^2+y^2+4xy+4x+32y+8=0$ (B) $4x^2+y^2+4xy+4x+32y+16=0$  

(C) $4x^2-y^2+4xy+4x+32y+12=0$ (D) $4x^2+2y^2+4xy+4x-32y+20=0$ 

Ans: B

6. The foci of an ellipse are (0, ±2) and its eccentricity is \frac{1}{\sqrt{2}} . then its equation is

(A) $\frac{x^2}{2}+\frac{y^2}{8}=1$ (B) $\frac{x^2}{4}+\frac{y^2}{6}=1$

(C)\$frac{x}^{2}4+\frac{y^2}{3}=1$ (D) $\frac{x^2}{4}+\frac{y^2}{8}=1$

Ans:  D

7. The equation of the ellipse with foci at (4, 6) and (16, 6), and a semi-minor axis of length 4, is

(A) $\frac{(x-10{)}^2}{52}+\frac{(y-6{)}^2}{16}=1$

(B) $\frac{(x-8{)}^2}{52}+\frac{(y-6{)}^2}{16}=1$

(C) $\frac{(x-10{)}^2}{52}+\frac{(y-6{)}^2}{4}=1$

(D) $\frac{(x-10{)}^2}{52}+\frac{(y-8{)}^2}{16}=1$

Ans:  A

8. Find the equation to the hyperbola, the distance between whose foci is 16 and whose eccentricity is $\sqrt{2}$ .

Ans: $x^2-y^2=32$

9. The eccentricity of the hyperbola $4x^2-9y^2-8x=32$ is -

(A) $\frac{\sqrt{5}}{3}$ (B) $\frac{\sqrt{13}}{3}$

(C) $\frac{\sqrt{13}}{2}$ (D)$\frac{3}{2}$

Ans: B