Coordinate Geometry Previous Year Questions with Solutions

Coordinate Geometry Previous Year Questions typically cover topics like the distance formula, midpoint, slope, equations of lines, and conic sections (circle, parabola, ellipse, hyperbola). Examples include finding the distance between two points, proving points are collinear, determining the equation of a line passing through two points, and analyzing the properties of conic sections. Solutions involve applying formulas like distance _{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}} , midpoint _{\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)}, and using standard equations for circles and parabolas. Practicing these questions helps in mastering key concepts and improving problem-solving skills.n

Coordinate Geometry Previous Year Questions for JEE with Solutions

JEE questions in Coordinate Geometry often test concepts related to straight lines, conic sections (circle, parabola, ellipse, hyperbola), and their properties. Some common types of problems include:

1. Equations of Lines: Questions on finding equations of lines passing through two points, lines parallel/perpendicular to a given line, or distance from a point to a line.
2. Circle and Parabola: Problems related to tangents, normals, and equations of circles and parabolas, including their focus, directrix, and eccentricity.
3. Ellipse and Hyperbola: Questions on standard forms, foci, asymptotes, and properties of ellipses and hyperbolas.

These questions are aimed at testing deeper understanding and application of the principles of Coordinate Geometry.

Note: In the JEE Main Mathematics exam, you can generally expect 4 to 6 questions from the Coordinate Geometry chapter.

1.0Key Concepts to Remember

Here are the Key Concepts to remember in Coordinate Geometry for IIT-JEE preparation:

1. Straight Lines:

• Slope of a Line: The slope (m) of a line passing through two points $(x_1,y_1)and(x_2,y_2)$ is given by:

$m=\frac{y_2-y_1}{x_2-x_1}$

• Equation of a Line: Various forms of the equation of a line:
• Point-Slope Form: $y-y_1=m(x-x_1)$
• Slope-Intercept Form: y = mx + c
• General Form: Ax + By + C = 0
• Distance from a Point to a Line:

$\text{ Distance }=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$ 

• Angle Between Two Lines:

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

2. Conic Sections:

• Circle: Standard equation:

$(x-h{)}^2+(y-k{)}^2=r^2$ 

where (h, k) is the center and r is the radius.

• Parabola: Standard equation:

$y^2=4ax\text{ or }{x}^2=4ay$

Focus and directrix are essential concepts for parabolas.

• Ellipse: Standard equation:

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

where a > b (major and minor axes).

• Hyperbola: Standard equation:

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

Asymptotes and foci are critical elements of hyperbolas.

3. Important Points and Properties:

• Midpoint Formula:

$\text{ Midpoint of }\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

• Collinearity of Points: Points $P_1(x_1,y_1),P_2(x_2,y_2),and{P}_3(x_3,y_3)$ are collinear if the area of the triangle they form is zero.$\text{ Area }=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=0$
• Section Formula: If a point P(x, y) divides a line segment joining $A(x_1,y_1)andB(x_2,y_2)$ in the ratio m:n, then:

$x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n}$

4. General Applications:

• Equation of a Line Parallel or Perpendicular to Another Line: Use slope for parallelism and negative reciprocal for perpendicular lines.
• Locus Problems: Deriving equations of loci (e.g., the set of points equidistant from a fixed point is a circle).
• Tangents and Normals: Understanding tangents to conic sections (circle, parabola, ellipse, hyperbola) and their equations.

These concepts are foundational for solving problems in Coordinate Geometry for JEE. Practice applying these principles in different scenarios to strengthen your understanding and problem-solving skills.

2.0JEE Mains Past Year Questions with Solutions on Coordinate Geometry

JEE Mains past year questions on Coordinate Geometry cover topics like 

1. Straight lines
2. Circles
3. Parabola
4. Ellipse
5. Hyperbola

3.0PYQs from Straight line

1. Consider a triangle ABC having the vertices A(1, 2), B(α, β) and C(γ, δ) and angles $\angle \mathrm{ABC}=\frac{\pi }{6}$  and $\angle \mathrm{BAC}=\frac{2\pi }{3}$. If the points B and C lie on the line y = x + 4, then α² + γ² is equal to …..  [JEE (Main) 2024]

Ans. (14)

Sol.

Equation of line passes through point A(1, 2) which makes angle $\frac{\pi }{6}$ from y = x + 4 is

$y-2=\frac{1\pm \tan \frac{\pi }{6}}{1⊠\tan \frac{\pi }{6}}(x-1)$

$\begin{array}{rl}& \begin{array}{cc}\oplus & \Theta \\ y-2=(2+\sqrt{3})(x-1)& y-2=(2-\sqrt{3})(x-1)\\ \text{ solve with }y=x+4& \text{ solve with }y=x+4\\ x+2=(2+\sqrt{3})x-2-\sqrt{3}& x+2=(2-\sqrt{3})x-2+\sqrt{3}\\ x=\frac{4+\sqrt{3}}{1+\sqrt{3}}& x=\frac{4-\sqrt{3}}{1-\sqrt{3}}\end{array}\\ & y-2=\frac{\sqrt{3}\pm 1}{\sqrt{3}\otimes 1}(x-1)\\ & {\alpha }^2+{\gamma }^2={\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)}^2+{\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)}^2\\ & {\alpha }^2+{\gamma }^2=14\end{array}$

2. Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that ΔOPQ is an isosceles triangle and ∠POQ = 90°. If l = OP² + PQ² + QO², then the greatest integer less than or equal to l is: 

(1) 44 (2) 48 

(3) 46 (4) 42  [JEE (Main) 2024]

Ans. (3)

Sol.

3x + 4y = 12 

3(rcosθ) + 4(rsinθ) = 12

r(3cosθ + 4sinθ) = 12   ...(1)

3(–rsinθ) + 4(rcosθ) = 12

r(–3sinθ + 4cosθ) = 12  ...(2)

${\left(\frac{12}{r}\right)}^2+{\left(\frac{12}{r}\right)}^2=(3\cos \theta +4\sin \theta {)}^2+(-3\sin \theta +4\cos \theta {)}^2$

$\begin{array}{rl}& 2{\left(\frac{12}{r}\right)}^2=9+16\\ & \frac{2\times 144}{r^2}=25\Rightarrow 288=25r^2\\ & \Rightarrow \frac{288}{25}=r^2\\ & \Rightarrow \sqrt{2}\left(\frac{12}{5}\right)=r\\ & \lambda =0P^2+P{Q}^2+Q{0}^2\\ & \lambda =r^2+r^2+r^2(\cos \theta +\sin \theta {)}^2+r^2(\sin \theta +\cos \theta {)}^2\\ & =2r^2+r^2(1+\sin 2\theta +1-2\sin 2\theta )\\ & =2r^2+2r^2\\ & =4r^2\end{array}$

$\begin{array}{rl}& =4\left(\frac{288}{25}\right)=\frac{1152}{25}=46.08\\ & [\lambda ]=46\end{array}$

3. If P(6, 1) is the orthocentre of the triangle whose vertices are A(5, –2), B(8, 3) and C(h, k), then point C lies on the circle. [JEE (Main) 2024]

(1) x² + y² – 65 = 0 (2) x² + y² – 74 = 0

(3) x² + y² – 61 = 0 (4) x² + y² – 52 = 0

Ans. (1)

Sol.

Slope of AD = 3

Slope of BC = -1/3

Equation of BC = 3y + x – 17 = 0

Slope of BE = 1

Slope of AC = –1

Equation of AC is x + y – 3 = 0

Point C is (–4, 7)

4. If the line segment joining the points (5, 2) and (2, a) subtends an angle $\frac{\pi }{4}$ at the origin, then the absolute value of the product of all possible values of a is:

(1) 6 (2) 8

(3) 2 (4) 4 [JEE (Main) 2024]

Ans. (4)

Sol.

$\begin{array}{rl}& m_{OA}=\frac{2}{5}\\ & m_{OB}=\frac{a}{2}\\ & \tan \frac{\pi }{4}=\left|\frac{2}{5}-\frac{a}{2}\right|\\ & 1=\left|\frac{4-5a}{10+2a}\right|\\ & 4-5a=\pm (10+2a)\\ & 4-5a=10+2a\\ & \Rightarrow \mid 7a+6=0\\ & a=-\frac{6}{7}\\ & -\frac{6}{7}\times \frac{14}{3}=-44-5a=-10-2a\\ & 3a=14\\ & a=+\frac{14}{3}\\ & a\end{array}$

5. Let ABC be an isosceles triangle in which A is at (–1, 0), $\angle \mathrm{A}=\frac{2\pi }{3}$, AB = AC and B is on the positive x-axis. If $BC=4\sqrt{3}$ and the line BC intersects the line y = x + 3 at (α, β), then $\frac{{\beta }^4}{{\alpha }^2}$ is : [JEE (Main) 2024]

Ans. (36)

Sol.

$\begin{array}{rl}& \frac{c}{\sin 30^{\circ }}=\frac{4\sqrt{3}}{\sin 120^{\circ }}\text{ [By sine rule] }\\ & 2c=8\Rightarrow c=4\\ & AB=|(b+1)|=4\\ & b=3,m_{AB}=0\\ & m_{BC}=\frac{-1}{\sqrt{3}}\end{array}\begin{array}{rl}& \text{ BC:- }y=\frac{-1}{\sqrt{3}}(x-3)\\ & \sqrt{3}y+x=3\end{array}$

$\begin{array}{rl}& \text{ Point of intersection : }y=x+3,\sqrt{3}y+x=3\\ & (\sqrt{3+1})y=6\\ & y=\frac{6}{\sqrt{3}+1}\\ & x=\frac{6}{\sqrt{3}+1}-3\\ & =\frac{6-3\sqrt{3}-3}{\sqrt{3}+1}\\ & =3\frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3}{)}^2}\\ & \frac{{\beta }^4}{{\alpha }^2}=36\end{array}$

6. Let A(–2, –1), B(1, 0), $C(\alpha ,\beta )$ and $\mathrm{D}(\gamma ,\delta )$ be the vertices of a parallelogram ABCD. If the point C lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of $|\alpha +\beta +\gamma +\delta |$ is equal to ________. [JEE (Main) 2024]

Ans. (32)

Sol.-

$\begin{array}{rl}& \mathrm{P}\equiv \left(\frac{\alpha -2}{2},\frac{\beta -1}{2}\right)\equiv \left(\frac{\gamma +1}{2},\frac{\delta }{2}\right)\\ & \frac{\alpha -2}{2}=\frac{\gamma +1}{2}\text{ and }\frac{\beta -1}{2}=\frac{\delta }{2}\\ & \Rightarrow \alpha -\gamma =3\dots .(1),\beta -\delta =1\dots ..\end{array}$

Also, $(\gamma ,\delta )$ lies on 3x – 2y = 6

$3\gamma -2\delta =6\dots \dots (3)$

and $(\alpha ,\beta )$ lies on 2x – y = 5 

$\Rightarrow 2\alpha -\beta =5.\dots \dots \dots$_.

Solving (1), (2), (3), (4)

$\begin{array}{rl}& \alpha =-3,\beta =-11,\gamma =-6,\delta =-12\\ & |\alpha +\beta +\gamma +\delta |=32\end{array}$

7. Let $\left(5,\frac{\mathrm{a}}{4}\right)$, be the circumcenter of a triangle with vertices $\mathrm{A}(\mathrm{a},-2),\mathrm{B}(\mathrm{a},6)$ and $\mathrm{C}\left(\frac{\mathrm{a}}{4},-2\right)$. Let $\alpha$ denote the circumradius, denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha +\beta +\gamma$ is

(1) 60 (2) 53

(3) 62 (4) 30 [JEE (Main) 2024]

Ans. (2)

Sol.

$\begin{array}{rl}& \mathrm{A}(\mathrm{a},-2),\mathrm{B}(\mathrm{a},6),\mathrm{C}\left(\frac{\mathrm{a}}{4},-2\right),\mathrm{O}\left(5,\frac{\mathrm{a}}{4}\right)\\ & \mathrm{AO}=\mathrm{BO}\\ & (\mathrm{a}-5{)}^2+{\left(\frac{\mathrm{a}}{4}+2\right)}^2=(\mathrm{a}-5{)}^2+{\left(\frac{\mathrm{a}}{4}-6\right)}^2\\ & \mathrm{a}=8\\ & \mathrm{AB}=8,\mathrm{AC}=6,\mathrm{BC}=10\\ & \alpha =5,\beta =24,\gamma =24\end{array}$

4.0PYQs from Circles

1. Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :

(1) r = 1 (2) r² – 8r + 8 = 0

(3) 2r² – 4r + 1 = 0 (4) 2r² – 8r + 7 = 0 [JEE (Main) 2024]

Ans. (2)

Sol.

OF² = r²

(2 – r)² + (2 – r)² = r²

r² – 8r + 8 = 0

2. Let the maximum and minimum values of ${\left(\sqrt{8x-x^2-12}-4\right)}^2+(x-7{)}^2$, x ∈ R be M and m respectively. Then M² – m² is equal to _____. [JEE (Main) 2024]

Ans. (1600)

Sol.

$\begin{array}{rl}& (x-7{)}^2+(y-4{)}^2\\ & y=\sqrt{8x-x^2-12}\\ & y^2=-(x-4{)}^2+16-12\\ & (x-4{)}^2+y^2=4\end{array}$

m = 9

M = 41

M² – m² = 41² – 9² = 1600

3. A square is inscribed in the circle x² + y² – 10x – 6y + 30 = 0. One side of this square is parallel to y = x + 3. If (x_i, y_i) are the vertices of the square, then

$\sum \left(x_i^2+y_i^2\right)$

is equal to :

(1) 148 (2) 156

(3) 160 (4) 152 [JEE (Main) 2024]

Ans. (4)

Sol.

y = x + c & x + y + d = 0

$\begin{array}{ll}\left|\frac{5-3+c}{\sqrt{2}}\right|=\sqrt{2}& \left|\frac{8+d}{\sqrt{2}}\right|=\sqrt{2}\\ |c+2|=2& 8+d=\pm 2\\ c=0,-4& d=-10,-6\\ \text{ pts }(5,5),(3,3),(7,3),(5,1)& \\ \sum \left(x_i^2+y_1^2\right)=25+25+9+9+49+9+25+1& \\ =152& \\ \text{ Option (4) }& \end{array}$

4. If the image of the point (-4, 5) in the line x + 2y = 2 lies on the circle (x + 4)² + (y - 3)² = r², then r is equal to:

(1) 1 (2) 2

(3) 75 (4) 3 

[JEE (Main) 2024]

Ans. (2)

Sol. Image of point (–4, 5)

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-2\left(\frac{a{x}_1+b{y}_1+c}{a^2+b^2}\right)$

Line: x + 2y – 2 = 0

$\begin{array}{rl}& \frac{x+4}{1}=\frac{y-5}{2}=-2\left(\frac{-4+10-2}{1^2+2^2}\right)\\ & =\frac{-8}{5}\end{array}$

$\begin{array}{rl}& x=-4-\frac{8}{5}=-\frac{28}{5}\\ & y=-\frac{16}{5}+5=\frac{9}{5}\end{array}$

Point lies on circle (x + 4)² + (y – 3)² = r²

$\begin{array}{rl}& \frac{64}{25}+{\left(\frac{9}{5}-3\right)}^2=r^2\\ & \frac{100}{25}=r^2,r=2\end{array}$

5. Let the centre of a circle, passing through the point (0, 0), (1, 0) and touching the circle x² + y² = 9, be (h, k). Then for all possible values of the coordinates of the centre (h, k), 4(h² + k²) is equal to ________.

[JEE (Main) 2024]

Ans. (9)

Sol.

(x – h)² + (y – k)² = h² + k²

x² + y² – 2hx – 2ky = 0

∵ passes through (1, 0)

⇒ 1 + 0 – 2h = 0

⇒ h = 1/2

$\begin{array}{rl}& \because \mathrm{OC}=\frac{\mathrm{OP}}{2}\\ & \sqrt{{\left(\frac{1}{2}\right)}^2+{\mathrm{k}}^2}=\frac{3}{2}\\ & \frac{1}{4}+{\mathrm{k}}^2=\frac{9}{4}\\ & {\mathrm{k}}^2=2\\ & \mathrm{k}=\pm \sqrt{2}\end{array}$

∴ Possible coordinate of 

$\begin{array}{rl}& \mathrm{c}(\mathrm{h},\mathrm{k})\left(\frac{1}{2},\sqrt{2}\right)\left(\frac{1}{2},-\sqrt{2}\right)\\ & 4\left({\mathrm{h}}^2+{\mathrm{k}}^2\right)=4\left(\frac{1}{4}+2\right)=4\left(\frac{9}{4}\right)=9\end{array}$

6. Equation of two diameters of a circle are $2x-3y=5$ and $3x-4y=7$. The line joining the points  $\left(-\frac{22}{7},-4\right)\text{ and }\left(-\frac{1}{7},3\right)$ intersects the circle at only one point $P(\alpha ,\beta )\text{. Then }17\beta -\alpha$.  is equal to 

[JEE (Main) 2024]

Ans. (2)

Sol. Centre of circle is (1, −1)

Equation of AB is 7x – 3y + 10 = 0 …(i)

Equation of CP is 3x + 7y + 4 = 0 …(ii)

Solving (i) and (ii)

$\begin{array}{rl}& \alpha =\frac{-41}{29},\beta =\frac{1}{29}\\ & \therefore 17\beta -\alpha =2\end{array}$

7. Consider two circles C₁ : x² + y² = 25 and C₂ : (x – α)² + y² = 16, where α ∈ (5, 9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C₁ and C₂ be ${\sin }^{-1}\left(\frac{\sqrt{63}}{8}\right)$. If the length of the common chord of C₁ and C₂ is β, then the value of (αβ)² equals _____ . [JEE (Main) 2024]

Ans. (1575)

Sol.

$C_1:x^2+y^2=25C_2:(x-\alpha {)}^2+y^2=16$

5 < α < 9

$\begin{array}{rl}& \theta ={\sin }^{-1}\left(\frac{\sqrt{63}}{8}\right)\\ & \sin \theta =\frac{\sqrt{63}}{8}\\ & \text{ Area of }△\mathrm{OAP}=\frac{1}{2}\times \alpha \left(\frac{\beta }{2}\right)=\frac{1}{2}\times 5\times 4\sin \theta \\ & \Rightarrow \alpha \beta =40\times \frac{\sqrt{63}}{8}\\ & \alpha \beta =5\times \sqrt{63}\\ & (\alpha \beta {)}^2=25\times 63=1575\end{array}$

8. Let a variable line passing through the centre of the circle x² + y² – 16x – 4y = 0, meet the positive co-ordinate axes at the point A and B. Then the minimum value of OA + OB, where O is the origin, is equal to 

(1) 12

(2) 18

(3) 20

(4) 24 [JEE (Main) 2024]

Ans. (2)

Sol.-

$\begin{array}{rl}& (\mathrm{y}-2)=\mathrm{m}(\mathrm{x}-8)\\ & \Rightarrow \mathrm{x}\text{-intercept }\\ & \Rightarrow \left(\frac{-2}{\mathrm{m}}+8\right)\\ & \Rightarrow \mathrm{y}\text{-intercept }\\ & \Rightarrow (-8\mathrm{m}+2)\\ & \Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{m}}+8-8\mathrm{m}+2\\ & {\mathrm{f}}^{\prime }(\mathrm{m})=\frac{2}{{\mathrm{m}}^2}-8=0\\ & \Rightarrow {\mathrm{m}}^2=\frac{1}{4}\\ & \Rightarrow \mathrm{m}=\frac{-1}{2}\\ & \Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18\\ & \Rightarrow \mathrm{Minimum}=18\end{array}$

5.0PYQs from Parabola

1. Let the length of the focal chord PQ of the parabola y² = 12x be 15 units. If the distance of PQ from the origin is p, then 10p² is equal to ____

[JEE (Main) 2024]

Ans. (72)

Sol.

length of focal chord 

$\begin{array}{rl}& =4a{\mathrm{cosec}}^2\theta =15\\ & 12{\mathrm{cosec}}^2\theta =15\\ & {\sin }^2\theta =\frac{4}{5}\\ & {\tan }^2\theta =4\\ & \tan \theta =2\\ & \text{ equation }\frac{y-0}{x-3}=2\\ & \mathrm{y}=2\mathrm{x}-6\\ & 2x-y-6=0\\ & P=\frac{6}{\sqrt{5}}\\ & 10{\mathrm{p}}^2=10\cdot \frac{36}{5}=72\end{array}$

2. Consider the circle C : x² + y² = 4 and the parabola P : y² = 8x. If the set of all values of α, for which three chords of the circle C on three distinct lines passing through the point (α, 0) are bisected by the parabola P is the interval (p, q), then (2q – p)² is equal to _____.

[JEE (Main) 2024]

Ans. (80)

Sol.

$\begin{array}{rl}& T=S_1\\ & x_1+y_1=x_1^2+y_1^2\\ & \alpha x_1=x_1^2+y_1^2\\ & \alpha \left(2t^2\right)=4t^4+16t^2\\ & \alpha =2t^2+8\\ & \frac{\alpha -8}{2}=t^2\\ & \text{ Also, }4t^4+16t^2-4<0\\ & t^2=-2+\sqrt{5}\\ & \alpha =4+2\sqrt{5}\\ & \therefore \alpha \in (8,4+2\sqrt{5})\\ & \therefore (2q-p{)}^2=80\end{array}$

3. Let A, B and C be three points on the parabola y² = 6x and let the line segment AB meet the line L through C parallel to the x-axis at the point D. Let M and N respectively be the feet of the perpendiculars from A and B on L. 

Then ${\left(\frac{\mathrm{AM}\cdot \mathrm{BN}}{\mathrm{CD}}\right)}^2$ is equal to _______.

[JEE (Main) 2024]

Ans. (36)

Sol.

$\begin{array}{rl}& m_{AB}=m_{AD}\\ & \Rightarrow \frac{2}{t_1+t_2}=\frac{2a\left(t_1-t_3\right)}{a{t}_1^2-\alpha }\\ & \Rightarrow a{t}_1^2-\alpha =a\left(t_1^2-t_1{t}_3+t_1{t}_2-t_2{t}_3\right\}\\ & \Rightarrow \alpha =a\left(t_1{t}_3+t_2{t}_3-t_1{t}_2\right)\\ & AM=\left|2a\left(t_1-t_3\right)\right|,BN=\left|2a\left(t_2-t_3\right)\right|,CD=\left|a{t}_3^2-\alpha \right|\\ & CD=\left|a{t}_3^2-a\left(t_1{t}_3+t_2{t}_3-t_1{t}_2\right)\right|\\ & =a\left|t_3^2-t_1{t}_3-t_2{t}_3+t_1{t}_2\right|\\ & =a\left|t_3\left(t_3-t_1\right)-t_2\left(t_3-t_1\right)\right|\\ & CD=a\left|\left(t_3-t_2\right)\left(t_3-t_1\right)\right|\\ & {\left(\frac{AM\cdot BN}{CD}\right)}^2={\left\{\frac{2a\left(t_1-t_3\right)\cdot 2a\left(t_2-t_3\right)}{a\left(t_3-t_2\right)\left(t_3-t_1\right)}\right\}}^2\\ & 16a^2=16\times \frac{9}{4}=36\end{array}$

4. Let the line L : $\sqrt{2}x+y=\alpha$ pass through the point of the intersection P (in the first quadrant) of the circle x² + y² = 3 and the parabola x² = 2y. Let the line L touch two circles C₁ and C₂ of equal radius $2\sqrt{3}$. If the centres Q₁ and Q₂ of the circles C₁ and C₂ lie on the y-axis, then the square of the area of the triangle PQ₁Q₂ is equal to ___________.

[JEE (Main) 2024]

Ans. (72)

Sol.

$\begin{array}{rl}& {\mathrm{x}}^2+{\mathrm{y}}^2=3\text{ and }{\mathrm{x}}^2=2\mathrm{y}\\ & {\mathrm{y}}^2+2\mathrm{y}-3=0\Rightarrow (\mathrm{y}+3)(\mathrm{y}-1)=0\\ & \mathrm{y}=-3\text{ or }\mathrm{y}=1\\ & \mathrm{y}=1\mathrm{x}=\sqrt{2}\Rightarrow P(\sqrt{2},1)\end{array}pliesontheline\begin{array}{rl}& \sqrt{2}x+y=\alpha \\ & \sqrt{2}(\sqrt{2})+1=\alpha \\ & \alpha =3\end{array}$

For circle C₁

Q₁ lies on y axis

Let Q₁ $(0,\alpha )$ coordinates

R₁ = $2\sqrt{3}$ (Given

Line L act as tangent

Apply P = r (condition of tangency)

$\begin{array}{rl}& \Rightarrow \left|\frac{\alpha -3}{\sqrt{3}}\right|=2\sqrt{3}\\ & \Rightarrow |\alpha -3|=6\\ & \alpha -3=6\text{ or }\alpha -3=-6\\ & \Rightarrow \alpha =9\\ & ⊠P{Q}_1{Q}_2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& 9& 1\\ 0& -3& 1\end{array}\right|\\ & =\frac{1}{2}(\sqrt{2}(12))=6\sqrt{2}\\ & {\left(⊠P{Q}_1{Q}_2\right)}^2=72\end{array}$

5. Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse $\mathrm{E}:\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\mathrm{a}>\mathrm{b}$ of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola P. Then the square of the length of the latus rectum of E, is

(1) _385/8

(2) _347/8

(3) _512/25

(4) _656/25

[JEE (Main) 2024]

Ans. (4)

Sol.-

Slope of axis = 1/2

$\begin{array}{rl}& y-3=\frac{1}{2}(x-2)\\ & \begin{array}{rl}& \Rightarrow 2y-6=x-2\\ & \Rightarrow 2y-x-4=0\\ & 2x+y-6=0\\ & 4x+2y-12=0\\ & \alpha +1.6=4\Rightarrow \alpha =2.4\\ & \beta +2.8=6\Rightarrow \beta =3.2\end{array}\end{array}$

Ellipse passes through (2.4, 3.2) 

$\begin{array}{rl}& \Rightarrow \frac{{\left(\frac{24}{10}\right)}^2}{{\mathrm{a}}^2}+\frac{{\left(\frac{32}{10}\right)}^2}{{\mathrm{b}}^2}=1\\ & \text{ Also }1-\frac{b^2}{a^2}=\frac{1}{2}=\frac{b^2}{a^2}=\frac{1}{2}\\ & \Rightarrow {\mathrm{a}}^2=2{\mathrm{b}}^2\end{array}$

Put in (1) 

$\begin{array}{rl}& \Rightarrow {\mathrm{b}}^2=\frac{328}{25}\\ & \Rightarrow {\left(\frac{2{\mathrm{b}}^2}{\mathrm{a}}\right)}^2=\frac{4{\mathrm{b}}^2}{{\mathrm{a}}^2}\times {\mathrm{b}}^2=4\times \frac{1}{2}\times \frac{328}{25}=\frac{656}{25}\end{array}$

6.0PYQs from Ellipse

1. The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1,\frac{2}{5}\right)$, is equal to :

(1) $\frac{\sqrt{1691}}{5}$ (2) $\frac{\sqrt{2009}}{5}$

(3) $\frac{\sqrt{1741}}{5}$ (4) $\frac{\sqrt{1541}}{5}$ [JEE (Main) 2024]

Ans. (1)

Sol. Equation of chord with given middle point.

$\begin{array}{rl}& T=S_1\\ & \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100}\\ & \frac{8x+5y}{200}=\frac{8+2}{200}\\ & y=\frac{10-8x}{5}\end{array}$

$\begin{array}{rl}& \frac{x^2}{25}+\frac{(10-8x{)}^2}{400}=1\\ & \text{ (put in original equation) }\\ & \frac{16x^2+100+64x^2-160x}{400}=1\\ & 4x^2-8x-15=0\\ & \mathrm{x}=\frac{8\pm \sqrt{304}}{8}\\ & {\mathrm{x}}_1=\frac{8+\sqrt{304}}{8};{\mathrm{x}}_2=\frac{8-\sqrt{304}}{8}\end{array}$

$\begin{array}{rl}& \text{ Similarly, }\\ & \mathrm{y}=\frac{10-18\pm \sqrt{304}}{5}=\frac{2\pm \sqrt{304}}{5}\\ & \begin{array}{rl}& {\mathrm{y}}_1=\frac{2-\sqrt{304}}{5};{\mathrm{y}}_2=\frac{2+\sqrt{304}}{5}\\ & \text{ Distance }=\sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}\\ & =\sqrt{\frac{4\times 304}{64}+\frac{4\times 304}{25}}=\frac{\sqrt{1691}}{5}\end{array}\end{array}$

7.0PYQs from Hyperbola

1. Consider a hyperbola H having a centre at the origin and foci and the x-axis. Let C₁ be the circle touching the hyperbola H and having the centre at the origin. Let C₂ be the circle touching the hyperbola H at its vertex and having the centre at one of its foci. If areas (in sq. units) of C₁ and C₂ are 36π and 4π, respectively, then the length (in units) of latus rectum of H is  

(1) _28/3 (2) _14/3

(3) _10/3 (4) _11/3 [JEE (Main) 2024]

Ans. (1)

Sol. Let H :

$\begin{array}{rl}& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\left(b^2=a^2\left(e^2-1\right)\right)\\ & \therefore {\text{ eq }}^n\text{ of }{C}_1=x^2+y^2=a^2\\ & \text{ Ar. }=36\pi \\ & \pi a^2=36\pi \\ & a=6\\ & \text{ Now radius of }{C}_2\text{ can be }a(e-1)\text{ or }a(e+1)\\ & \text{ for }r=a(e-1)\text{ for }r=a(e+1)\\ & \text{ Ar. }=4\pi \pi r^2=4\pi \end{array}$

$\begin{array}{ll}\pi a^2(e-1{)}^2=4\pi & a^2(e+1{)}^2=4\\ 36\pi (e-1{)}^2=4\pi & 36(e+1{)}^2=4\\ e-1=\frac{1}{3}& e+1=\frac{1}{3}\\ e=\frac{4}{3}& -\frac{2}{3}\\ \therefore b^2=36\left(\frac{16}{9}-1\right)=28& \\ \therefore LR=\frac{2b^2}{a}=\frac{2\times 28}{6}=\frac{28}{3}& \end{array}$

2. Let S be the focus of the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$ , on the positive x-axis. Let C be the circle with its centre at A$(\sqrt{6},\sqrt{5})$ and passing through the point S. if O is the origin and SAB is a diameter of C then the square of the area of the triangle OSB is equal to - [JEE (Main) 2024]

Ans. ($\sqrt{40}$)

Sol.

$\text{ Area }=\frac{1}{2}(\mathrm{OS})\mathrm{h}=\frac{1}{2}\sqrt{8}2\sqrt{5}=\sqrt{40}$

3. If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is _15/8 times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2},\frac{14}{3}\sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to 

$(1)7\sqrt{\frac{2}{5}}-\frac{8}{3}(2)14\sqrt{\frac{2}{5}}-\frac{4}{3}(3)14\sqrt{\frac{2}{5}}-\frac{16}{3}(4)7\sqrt{\frac{2}{5}}+\frac{8}{3}$ [JEE (Main) 2024]

Ans. (1)

Sol.

$\begin{array}{rl}& \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{25}=1\\ & \mathrm{a}=3,\mathrm{b}=5\\ & \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\therefore \text{ foci }=(0,\pm \mathrm{be})=(0,\pm 4)\\ & \therefore {\mathrm{e}}_{\mathrm{H}}=\frac{4}{5}\times \frac{15}{8}=\frac{3}{2}\end{array}$

Let equation hyperbola

$\begin{array}{rl}& \frac{x^2}{A^2}-\frac{y^2}{B^2}=-1\\ & \therefore \mathrm{B}\cdot {\mathrm{e}}_{\mathrm{H}}=4\therefore \mathrm{B}=\frac{8}{3}\end{array}\begin{array}{rl}& \begin{array}{rl}& \therefore A^2=B^2\left(e_H^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right)\therefore A^2=\frac{80}{9}\\ & \therefore \frac{x^2}{\frac{80}{9}}-\frac{y^2}{\frac{64}{9}}=-1\\ & \text{ Directrix : }y=\pm \frac{B}{e_H}=\pm \frac{16}{9}\end{array}\\ & PS=\mathrm{e}\cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3}\cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right|\\ & =7\sqrt{\frac{2}{5}}-\frac{8}{3}\end{array}$

4. Let the foci and length of the latus rectum of an ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b\text{ be }(\pm 5,0)$, respectively. Then, the square of the eccentricity of the hyperbola  equals [JEE (Main) 2024]

Ans. (51)

Sol.

$\begin{array}{rl}& \text{ focii }\equiv (\pm 5,0);\frac{2b^2}{a}=\sqrt{50}\\ & e=5b^2=\frac{5\sqrt{2}a}{2}\\ & b^2=a^2\left(1-e^2\right)=\frac{5\sqrt{2}a}{2}\end{array}$

$\begin{array}{rl}& \Rightarrow a\left(1-e^2\right)=\frac{5\sqrt{2}}{2}\\ & \Rightarrow \frac{5}{e}\left(1-e^2\right)=\frac{5\sqrt{2}}{2}\\ & \Rightarrow \sqrt{2}-\sqrt{2}{e}^2=e\\ & \Rightarrow \sqrt{2}{e}^2+e-\sqrt{2}=0\\ & \Rightarrow \sqrt{2}{e}^2+2e-e-\sqrt{2}=0\\ & \Rightarrow \sqrt{2}e(e+\sqrt{2})-1(1+\sqrt{2})=0\\ & \Rightarrow (e+\sqrt{2})(\sqrt{2}e-1)=0\\ & \therefore e\ne -\sqrt{2};e=\frac{1}{\sqrt{2}}\\ & \frac{x^2}{b^2}-\frac{y^2}{a^2{b}^2}=1a=5\sqrt{2}\\ & a^2{b}^2=b^2\left(e_1^2-1\right)\Rightarrow e_1^2=51,b=5\end{array}$