Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 11 :

Consider ellipses $E_{k} : k x^{2} + k^{2} y^{2} = 1, k = 1, 2, \dots, 20 .$ Let $C_{k}$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $E_{k}$ . If $r_{k}$ is the radius of the circle $C_{k}$ , then the value of $\sum_{k = 1}^{20} \frac{1}{r_{k}^{2}}$ is [2023]

3080
3210
3320
2870

1

2

3

4

(1)

$k x^{2} + k^{2} y^{2} = 1$

$\frac{x^{2}}{1 / k} + \frac{y^{2}}{1 / k^{2}} = 1$

Now, equation of line $A B$ is

$\frac{x}{1 / \sqrt{k}} + \frac{y}{1 / k} = 1 \Rightarrow \sqrt{k} x + k y = 1$

$r_{k} = ⊥ r$ distance of (0, 0) from line AB

$r_{k} =$ $| \frac{(0 + 0 - 1)}{\sqrt{k + k^{2}}} |$ $= \frac{1}{\sqrt{k + k^{2}}}$

$\frac{1}{r_{k}^{2}} = k + k^{2}$ $\Rightarrow \sum_{k = 1}^{20} \frac{1}{r_{k}^{2}} = \sum_{k = 1}^{20} (k + k^{2}) = \sum_{k = 1}^{20} k + \sum_{k = 1}^{20} k^{2}$

$= \frac{20 \cdot 21}{2}$ $+ \frac{20 \cdot 21 \cdot 41}{6} = 210 + 10 \times 7 \times 41$

$= 210 + 2870 = 3080$

Q 12 :

If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $x^{2} + 4 y^{2} = 36$ is $r$ , then $12 r^{2}$ is equal to [2023]

72
69
115
92

1

2

3

4

(4)

Centre of the circle is (2, 0).

Ellipse : $x^{2} + 4 y^{2} = 36$

$\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1$ $\dots (i)$

The equation of the circle is, ${(x - 2)}^{2} + y^{2} = r^{2}$ $\dots (i i)$

Solving (i) and (ii), we get

${(x - 2)}^{2} + \frac{36 - x^{2}}{4} = r^{2}$ $\Rightarrow x^{2} + 4 - 4 x + \frac{36 - x^{2}}{4} = r^{2}$

$\Rightarrow 4 x^{2} + 16 - 16 x + 36 - x^{2} = 4 r^{2}$

$\Rightarrow 3 x^{2} - 16 x + 52 - 4 r^{2} = 0$

Discriminant = 0

$∴$ ${(16)}^{2} - 4 \times 3 \times (52 - 4 r^{2}) = 0$

$\Rightarrow 52 - 4 r^{2} = \frac{256}{12}$ $\Rightarrow 4 r^{2} = 52 - \frac{256}{12} = \frac{368}{12}$

$\Rightarrow 12 r^{2} = \frac{368}{4} = 92$

Q 13 :

Let $P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}})$ , Q, R and S be four points on the ellipse $9 x^{2} + 4 y^{2} = 36$ . Let PQ and RS be mutually perpendicular and pass through the origin. If $\frac{1}{{(P Q)}^{2}} + \frac{1}{{(R S)}^{2}} = \frac{p}{q},$ where $p$ and $q$ are coprime, then $p + q$ is equal to [2023]

157
143
137
147

1

2

3

4

(1)

Q 14 :

Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{4} = 1$ meet the $y$ -axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line $x = 2 \sqrt{5}$ intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point $(α, β)$ , then $α^{2} - β^{2}$ is equal to [2023]

$\frac{314}{5}$
61
$\frac{304}{5}$
60

1

2

3

4

(3)

We have, $\frac{x^{2}}{36} + \frac{y^{2}}{4} = 1$ ...(i)

$∴$ The equation of tangent at point $(3 \sqrt{3}, 1)$ on the given ellipse is given by

$y - 1 = - 3 \sqrt{3} (\frac{4}{36}) (x - 3 \sqrt{3})$ $[∵ y - y_{1} = \frac{- x_{1}}{y_{1}} (\frac{b^{2}}{a^{2}}) (x - x_{1})]$

$\Rightarrow y - 1 = \frac{- x}{\sqrt{3}} + 3$ ...(i)

$\Rightarrow \sqrt{3} y - \sqrt{3} = - x + 3 \sqrt{3} \Rightarrow x + \sqrt{3} y = 4 \sqrt{3}$ ...(ii)

The equation of normal at point $(3 \sqrt{3}, 1)$ on the given ellipse

$y - 1 = \frac{1}{3 \sqrt{3}} (\frac{36}{4}) (x - 3 \sqrt{3})$ $[∵ y - y_{1} = (\frac{y_{1}}{x_{1}}) (\frac{a^{2}}{b^{2}}) (x - x_{1})]$

$\Rightarrow y - 1 = \sqrt{3} (x - 3 \sqrt{3}) \Rightarrow y - 1 = \sqrt{3} x - 9 \Rightarrow \sqrt{3} x - y = 8$

Tangent meets $y$ –axis at $y$ = 4 and normal meets $y$ –axis at $y$ = - 8.

$∴$ $A \equiv (0, 4)$ and $B \equiv (0, - 8)$

$∴$ $Equation of circle with diameter end points A and B is x^{2} + (y - 4) (y + 8) = 0$

$[∵ Equation of circle with diameter end points (x_{1}, y_{1}) is (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0]$

$\Rightarrow x^{2} + y^{2} + 4 y - 32 = 0$

$∴$ $Centre of circle = (0, - 2)$

Equation of line passing through centre of circle, $y = - 2$ .

Now, the equation of chord of contact PQ of two tangents drawn from the point $(α, β)$ is given by

$α x + β y + 2 (y + β) - 32 = 0$

Since $(2 \sqrt{5}, 0)$ lie on the chord of contact PQ

$\Rightarrow α (2 \sqrt{5}) + 2 (0 - 2) - 32 = 0 [∵ β = - 2]$

$\Rightarrow α (2 \sqrt{5}) = 36 \Rightarrow α = \frac{36}{2 \sqrt{5}}$

$∴$ $α^{2} - β^{2} = {(\frac{36}{2 \sqrt{5}})}^{2} - {(- 2)}^{2} = \frac{1296}{20} - 4 = \frac{1216}{20} = \frac{304}{5}$

Q 15 :

If the maximum distance of normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1, b < 2,$ from the origin is 1, then the eccentricity of the ellipse is: [2023]

$\frac{\sqrt{3}}{4}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$

1

2

3

4

(2)

We have, $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{b^{2}} = 1; b < 2$

Equation of normal at $(2 \cos θ, b \sin θ)$ is

$2 x \sec θ - b y c o s e c θ = 4 - b^{2} .$

According to the question, $| \frac{4 - b^{2}}{\sqrt{4 \sec^{2} θ + b^{2} {cosec}^{2} θ}} |_{\max}$ $= 1$

For maximum distance, $4 \sec^{2} θ + b^{2} {cosec}^{2} θ$ should be minimum.

$∴$ $\min (a^{2} \sec^{2} θ + b^{2} {cosec}^{2} θ) = {(a + b)}^{2}$

$∴$ $\Rightarrow 1 = \frac{4 - b^{2}}{\sqrt{{(2 + b)}^{2}}} \Rightarrow 1 = \frac{4 - b^{2}}{2 + b}$

$\Rightarrow 2 + b = 4 - b^{2} \Rightarrow b^{2} + b - 2 = 0$

$\Rightarrow b^{2} + 2 b - b - 2 = 0 \Rightarrow b (b + 2) - 1 (b + 2) = 0$

$\Rightarrow (b + 2) (b - 1) = 0 (∵ b < 2)$

$\Rightarrow b = 1$

Now, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

Q 16 :

Let an ellipse with centre (1, 0) and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$ -axis. If its minor axis subtends an angle $60 °$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________. [2023]

(9)

Length of latus rectum $= \frac{1}{2}$

and eccentricity $(a e) = \sin θ$ , where $θ$ is the angle subtended by minor axis at focus.

$\Rightarrow$ $a e = \sin 60 ° = \frac{\sqrt{3}}{2}$

Now, $\frac{2 b^{2}}{a} = \frac{1}{2} \Rightarrow b^{2} = \frac{a}{4} and a e = \frac{\sqrt{3}}{2}$

Now, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = a^{2} - {(a e)}^{2} = a^{2} - \frac{3}{4}$

$\Rightarrow$ $\frac{a}{4} = a^{2} - \frac{3}{4} \Rightarrow 4 a^{2} - a - 3 = 0 \Rightarrow (4 a + 3) (a - 1) = 0$

$\Rightarrow a = 1 \Rightarrow b = \frac{1}{2}$

$∴$ ${(2 a + 2 b)}^{2} = {(2 + 2 \times \frac{1}{2})}^{2} = {(3)}^{2} = 9$

Q 17 :

The line $x = 8$ is the directrix of the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the corresponding focus $(2, 0)$ . If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4 \sqrt{3})$ and intersects the $x$ -axis at $Q$ , then ${(3 P Q)}^{2}$ is equal to __________ . [2023]

(39)

Given, the line $x = 8$ is the directrix of the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{a}{e} = 8$ ...(i) and focus, $a e = 2$ ...(ii)

From (i) and (ii)

$8 e = \frac{2}{e} \Rightarrow e^{2} = \frac{1}{4}$ $∴$ $e = \frac{1}{2}$

and $a = 4, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 16 (\frac{3}{4}) = 12$

Now, equation of tangent at $P (4 \cos θ, 2 \sqrt{3} \sin θ)$ is

$\frac{x \cos θ}{4} + \frac{y \sin θ}{2 \sqrt{3}} = 1$ and it passes through $(0, 4 \sqrt{3})$ , so it satisfies the tangent equation.

$0 + \frac{4 \sqrt{3} \sin θ}{2 \sqrt{3}} = 1 \Rightarrow \sin θ = \frac{1}{2}$ $∴$ $θ = 30 °$

$∴$ $Point P = (2 \sqrt{3}, \sqrt{3}) and Q (\frac{8}{\sqrt{3}}, 0)$

$P Q^{2} = {(2 \sqrt{3} - \frac{8}{\sqrt{3}})}^{2} + {(\sqrt{3} - 0)}^{2} = \frac{13}{3}$

$∴$ ${(3 P Q)}^{2} = 9 \times P Q^{2} = 9 \times \frac{13}{3} = 39$

Q 18 :

Let $C$ be the largest circle centred at $(2, 0)$ and inscribed in the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If $(1, α)$ lies on $C$ , then $10 α^{2}$ is equal to _______ . [2023]

(118)

Let P be the point where ellipse and circle touch each other.

Let $P$ be $(6 \cos θ, 4 \sin θ)$

Equation of tangent to ellipse at point P is

$\frac{x \cos θ}{6} + \frac{y \sin θ}{4} = 1$ $\Rightarrow y = \frac{4}{\sin θ} (\frac{- x \cos θ}{6} + 1)$

$\Rightarrow y = - \frac{2 \cos θ}{3 \sin θ} x + \frac{4}{\sin θ}$

Let $m_{1} = \frac{- 2 \cos θ}{3 \sin θ}$

Slope of normal to circle at $P = m_{2}$ .

$m_{2} = \frac{(4 \sin θ - 0)}{(6 \cos θ - 0)}$ $∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \sin θ = \frac{4}{5}$

So, $P \equiv (\frac{18}{5}, \frac{16}{5})$

$r^{2} = {(\frac{18}{5} - 2)}^{2} + {(\frac{16}{5})}^{2} = {(1 - 2)}^{2} + α^{2} = \frac{64}{25} + \frac{256}{25} = 1 + α^{2}$

$α^{2} = \frac{320}{25} - 1 = \frac{64}{5} - 1 = \frac{59}{5}$

$\Rightarrow 10 α^{2} = 59 \times 2 = 118$

Q 19 :

Let a tangent to the curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _______ . [2023]

(7)

$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

Let $(4 \cos θ, 3 \sin θ)$ be the coordinates of the point at which the tangent is drawn.

Equation of tangent $\frac{x \cos θ}{4} + \frac{y \sin θ}{3} = 1$

Let $A = (4 \sec θ, 0)$

$B = (0, 3 c o s e c θ)$ $(Tangent intersects coordinate axes at A and B)$

$A B^{2} = 16 \sec^{2} θ + 9 c o s e c^{2} θ = 25 + 16 \tan^{2} θ + 9 \cot^{2} θ$

$A B^{2} \geq 25 + 2 \times \sqrt{16 \tan^{2} θ \times 9 \cot^{2} θ} = 49$

$A B \geq 7$

$∴$ $A B_{\min} = 7$

Q 20 :

Let the line $y - x = 1$ intersect the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is: [2026]

$π - \tan^{- 1} (\frac{1}{4})$
$\frac{π}{2} + \tan^{- 1} (\frac{1}{4})$
$\frac{π}{2} + 2 \tan^{- 1} (\frac{1}{4})$
$\frac{π}{2} - \tan^{- 1} (\frac{1}{4})$

1

2

3

4

(2)

$By solving line & equation of ellipse we get x = 0$

$& x = - \frac{4}{3}$

$∴$ $B (- \frac{4}{3}, - \frac{1}{3})$

$m_{O B} = \tan θ = \frac{1}{4}$

$∵$ $∠ A O B = \frac{π}{2} + θ = \frac{π}{2} + \tan^{- 1} (\frac{1}{4})$

Q 21 :

Let each of the two ellipses $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ and $E_{2} : \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1, (A < B)$ have eccentricity $\frac{4}{5}$ . Let the lengths of the latus recta of $E_{1}$ and $E_{2}$ be $l_{1}$ and $l_{2}$ , respectively, such that $2 l_{1}^{2} = 9 l_{2} .$ If the distance between the foci of $E_{1}$ is 8, then the distance between the foci of $E_{2}$ is [2026]

$\frac{8}{5}$
$\frac{96}{5}$
$\frac{16}{5}$
$\frac{32}{5}$

1

2

3

4

(4)

$2 a e = 8 \Rightarrow a = 5$

$b^{2} = a^{2} (1 - e^{2})$

$b^{2} = a^{2} \times \frac{9}{25} \Rightarrow b^{2} = 9$

$E_{1} : \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$

$ℓ_{1} : \frac{2 b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$

$A^{2} = B^{2} (1 - e^{2}) \Rightarrow A^{2} = \frac{9}{25} B^{2} \Rightarrow A = \frac{3}{5} B$

$2 ℓ^{2} = 9 ℓ_{2} \Rightarrow 2 {(\frac{18}{5})}^{2} = 9 ℓ_{2} \Rightarrow ℓ_{2} = \frac{4 \times 18}{25}$

$\frac{2 A^{2}}{B} = \frac{72}{25} \Rightarrow A^{2} = \frac{36}{25} B$

$\frac{9}{25} B^{2} = \frac{36}{25} B \Rightarrow B = 4$

$Distance between foci = 2 B e = 2 \times \frac{4}{5} \times 4 = \frac{32}{5}$

Q 22 :

An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

$\frac{16}{\sqrt{3}}$
$6 \sqrt{3}$
$4 \sqrt{3}$
6

1

2

3

4

(4)

$C A_{1} = a = 4$

$C F_{1} = a e = 2$

$e = \frac{1}{2}$

$L R = 2 e (\frac{a}{e} - a e)$

$= 2 \times \frac{1}{2} (\frac{4}{1 / 2} - 2)$

$= 6$

Q 23 :

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

256
516
276
496

1

2

3

4

(4)

$f (t) = \frac{- 3}{4} + 2 t - t^{2}$

$f (t) |_{maximum} = \frac{1}{4} = e \Rightarrow e^{2} = \frac{1}{16} \Rightarrow \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{16} . . . (1)$

$∵$ $\frac{2 b^{2}}{a} = 30 \Rightarrow b^{2} = 15 a . . . (2)$

$By (1) & (2)$

$16 (a^{2} - 15 a) = a^{2} \Rightarrow 15 a^{2} - 16 \times 15 a = 0$

$a = 16$

$b^{2} = 240$

$a^{2} + b^{2} = 256 + 240$

$= 496$

Q 24 :

If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $a b + 18 r^{2}$ is. [2026]

53
51
52
55

1

2

3

4

(4)

$By family of curve equation of circle will be$

$\Rightarrow S_{1} + λ S_{2} = 0$

$\Rightarrow (x^{2} + 2 y^{2} - 6 x - 12 y + 23) + λ (4 x^{2} + 2 y^{2} - 20 x - 12 y + 35) = 0$

$\Rightarrow for circle coeff of x^{2} = coeff of y^{2}$

$\Rightarrow λ = \frac{1}{2}$

$So equation of circle is$

$\Rightarrow x^{2} + y^{2} - \frac{16}{3} x - 6 y + \frac{27}{2} = 0$

$Centre (\frac{8}{3}, 3), Radius r = \sqrt{\frac{47}{18}}$

$∴$ $a b + 18 r^{2} = 8 + 47 = 55$

Q 25 :

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

$\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$
$\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{5}}$
$\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{11}}$
$\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{5}}$

1

2

3

4

(1)

$α x + 4 y - \sqrt{7} = 0 touches 3 x^{2} + 4 y^{2} = 1$

$∴$ $c^{2} = a^{2} m^{2} + b^{2}$

$\frac{7}{16} = \frac{1}{3} \times \frac{α^{2}}{16} + \frac{1}{4} \Rightarrow α = 3, - 3$

$Tangent is 3 x + 4 y - \sqrt{7} = 0$

Let the point of contact be $P (x_{1}, y_{1})$

$∴$ $Tangent is 3 x x_{1} + 4 y y_{1} = 1$

$∴$ $\frac{3 x_{1}}{3} = \frac{4 y_{1}}{4} = \frac{1}{\sqrt{7}}$ $∴$ $P (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$

$e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

$P S = e (P M)$

$= e (\frac{a}{e} - \frac{1}{\sqrt{7}})$

$= \frac{1}{2} (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{7}}$

$P S' = e (P M')$

$= \frac{1}{2} (\frac{a}{e} + \frac{1}{\sqrt{7}}) = \frac{1}{2} (\frac{1}{\sqrt{7}} + \frac{2}{\sqrt{3}})$

$= \frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$

Q 26 :

Let S and S′ be the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If ${(S P)}^{2} + {(S' P)}^{2} - S P \cdot S' P = 37,$ then $α^{2} + β^{2}$ is equal to: [2026]

13
15
17
11

1

2

3

4

(1)

$∵$ $P lies on ellipse \Rightarrow \frac{α^{2}}{25} + \frac{β^{2}}{9} = 1$

$∵$ $P S + P S' = 2 a \Rightarrow P S + P S' = 10$

$∴$ ${(P S)}^{2} + {(P S')}^{2} - P S \cdot P S' = 37$

${(P S + P S')}^{2} - 3 P S \cdot P S' = 37$

$100 - 3 P S \cdot P S' = 37$

$3 P S \cdot P S' = 63 \Rightarrow P S \cdot P S' = 21$

$∵$ $P S & P S' are (5 \pm \frac{4}{5} α)$

$∴$ $P S \cdot P S' = 25 - \frac{16}{25} α^{2} = 21$

$\frac{16}{25} α^{2} = 4$

$α = \frac{5}{2} \Rightarrow α^{2} = \frac{25}{4}$

$∴$ $β^{2} = \frac{27}{4}$

$∴$ $α^{2} + β^{2} = \frac{52}{4} = 13$

Q 27 :

Let the foci of the Ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and Hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the Hyperbola is:

$\frac{32}{9}$
$\frac{27}{10}$
$\frac{18}{5}$
$\frac{27}{4}$

1

2

3

4

(2)

$Foci of the Ellipse = (\pm 3, 0)$

$Foci of the Hyperbola = \pm \frac{1}{5} \sqrt{144 + α}$ $∴$ $α = 81$

$length of the latus rectum = \frac{2 b^{2}}{a} = \frac{27}{10}$

Q 28 :

An ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a < b)$ passes through the vertices of the hyperbola $H : \frac{x^{2}}{49} - \frac{y^{2}}{64} = - 1 .$ Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$ . Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$ . If $l$ is the length of the latus rectum of the ellipse $E$ , then the value of $113 l$ is equal to

1556
1552
1662
776

1

2

3

4

(2)

$Ellipse passes through (0, \pm 8)$

$b^{2} = 64 e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}, e_{H} = \frac{\sqrt{113}}{8}, λ = \frac{2 b^{2}}{a}$

Q 29 :

Let $P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}), Q, R$ and S be four points on ellipse $9 x^{2} + 4 y^{2} = 36$ . Let PQ and RS be mutually perpendicular chords and pass through the centre of ellipse, then the value of $[\frac{50}{P Q^{2}} + \frac{50}{R S^{2}}] =____$ where $[.]$ denotes GIF

4
5
6
7

1

2

3

4

(1)

Given ellipse:

$9 x^{2} + 4 y^{2} = 36$

and

$P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}})$

Since $P Q$ passes through the centre, $P Q$ is a diameter.

$O P^{2} = \frac{12}{7} + \frac{36}{7} = \frac{48}{7}$

$P Q^{2} = 4 O P^{2} = \frac{192}{7}$

Since $R S ⊥ P Q$ , slope of $O P = \sqrt{3}$ , so slope of $O R = - \frac{1}{\sqrt{3}}$ .

Put $y = - \frac{x}{\sqrt{3}}$ in the ellipse:

$9 x^{2} + \frac{4 x^{2}}{3} = 36$

$x^{2} = \frac{108}{31}$

Hence,

$O R^{2} = x^{2} + y^{2} = \frac{144}{31}$

$R S^{2} = 4 O R^{2} = \frac{576}{31}$

Therefore,

$\frac{50}{P Q^{2}} + \frac{50}{R S^{2}} = \frac{350}{192} + \frac{1550}{576} = \frac{325}{72} \approx 4.51$

Thus,

$[\frac{50}{P Q^{2}} + \frac{50}{R S^{2}}] = 4$

Q 30 :

If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to:

(75)

$We have x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$

$\Rightarrow \frac{{(x + 1)}^{2}}{λ + 5} + \frac{{(y + 1)}^{2}}{\frac{λ + 5}{4}} = 1$

$∵$ $\frac{2 b^{2}}{a} = 4$

$\frac{2 (λ + 5)}{4} = 4 \sqrt{λ + 5}$

$On solving \Rightarrow λ = 59$

$λ = - 5$

$l = 2 a = 2 \sqrt{λ + 5} = 2 \sqrt{65} = 16$

$\Rightarrow λ + l = 59 + 16 = 75$

Q 31 :

Match the items of Column-I with Column-II

Column-I Column-II

(I) If the coordinates of a point are $(4 \tan θ, 3 \sec θ)$ where
$θ (θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$ is parameter then the points lies on a conic whose eccentricity is P $\sqrt{3}$

(II) If an ellipse has the length of major axis 10 units and semi minor axis length 4 units, then this ellipse has eccentricity Q $\frac{4}{5}$

(III) If AB is double ordinate of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{9} = 1$ such that triangle OAB is an equilateral triangle of side '2' then eccentricity of hyperbola is (where O is centre of Hyperbola) R $\frac{5}{3}$

(IV) If the foci of ellipse $\frac{x^{2}}{K^{2} a^{2}} + \frac{y^{2}}{a^{2}} = 1$ and hyperbol $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$ coincide then K can be S $\frac{3}{5}$

T $\sqrt{\frac{13}{3}}$

I-R; II-S; III-T; IV-P
I-S; II-P; III-Q; IV-T
I-P; II-T; III-S; IV-Q
I-Q; II-S; III-P; IV-T

1

2

3

4

(1)

$I)$ $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ $e = \sqrt{\frac{16 + 9}{16}} = \frac{5}{4}$ $\frac{1}{e^{2}} + \frac{1}{e'^{2}} = 1$

$\frac{1}{e'^{2}} = 1 - \frac{16}{25} = \frac{9}{25}$ $e' = \frac{5}{3}$

$II)$ $2 a = 10 \Rightarrow a = 5$ $b^{2} = 16, b = 4$ $e = \sqrt{\frac{25 - 16}{25}}$

$III)$ $(a \sec θ, 3 \tan θ)$

$6 \tan θ = 2$ $\tan θ = \frac{1}{3}$ $\frac{1}{\sqrt{3}} = \frac{3 \tan θ}{a \sec θ}$

$a^{2} \sec^{2} θ = 27 \cdot \frac{1}{9}$

$a^{2} (1 + \frac{1}{9}) = 27 \cdot \frac{1}{9}$ $a^{2} (10) = 27$

$IV)$ $k^{2} a^{2} - a^{2} = a^{2} + a^{2}$

$k^{2} - 1 = 2 \Rightarrow k^{2} = 3 \Rightarrow \sqrt{3}$

Q 32 :

The length of focal chord AB of ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ is $(Given A = (\frac{8}{5}, \frac{3 \sqrt{3}}{5}))$

$\frac{4}{5}$
$\frac{16}{5}$
$\frac{32}{5}$
$\frac{64}{5}$

1

2

3

4

(2)

$∵$ $2 \cos θ_{1} = \frac{8}{5} \Rightarrow \cos θ_{1} = \frac{4}{5} \Rightarrow \tan \frac{θ_{1}}{2} = \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} = \frac{1}{3} and e = \frac{1}{2}$

$∵$ $\tan \frac{θ_{2}}{2} \times \frac{1}{3} = \frac{- \frac{1}{2}}{\frac{3}{2}} \Rightarrow \frac{θ_{2}}{2} = \frac{3 π}{4} \Rightarrow θ_{2} = \frac{3 π}{2}$

$B = (0, - \sqrt{3}) \Rightarrow A B = 2 + 2 - \frac{1}{2} \times \frac{8}{5} = 4 - \frac{4}{5} = \frac{16}{5}$

Q 33 :

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a^{2} = f o g (10), b^{2} = g o f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then $8 e^{2} + l^{2}$ is equal to

16
12
6
8

1

2

3

4

(4)

Given $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$

$a = f (g (10)) = f (\frac{10}{10 - 9}) = f (10) = 109$

$b = g (f (3)) = g (9 + 9) = g (18) = \frac{18}{9} = 2$

$E : \frac{x^{2}}{109} + \frac{y^{2}}{2} = 1; e^{2} = 1 - \frac{2}{109} = \frac{107}{109}; l = \frac{2 (2)}{\sqrt{109}} = \frac{4}{\sqrt{109}}; 8 e^{2} + l^{2} = \frac{8 (107)}{109} + \frac{16}{109} = 8$

Q 34 :

Let L be a common tangent line to the curves $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and $x^{2} + y^{2} = \frac{31}{4}$ . Then, the square of the slope of the line L is ________

(3)

Tangent to the curve $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is

$y^{2} = m x \pm \sqrt{a^{2} m^{2} + b^{2}}, y = m x \pm \sqrt{9 m^{2} + 4}$

And equation of tangent to the curve $x^{2} + y^{2} = \frac{31}{4}$ is

$y = m x \pm a \sqrt{1 + m^{2}}, y = m x \pm \sqrt{\frac{31}{4} (1 + m^{2})}$

For common tangent $9 m^{2} + 4 = \frac{31}{4} + \frac{31}{4} m^{2} \Rightarrow \frac{5}{4} m^{2} = \frac{15}{4} \Rightarrow m^{2} = 3$

Topic Question Set

Q 12 :

If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $x^{2} + 4 y^{2} = 36$ is $r$ , then $12 r^{2}$ is equal to [2023]

(1)

Q 15 :

If the maximum distance of normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1, b < 2,$ from the origin is 1, then the eccentricity of the ellipse is: [2023]

Q 16 :

Let an ellipse with centre (1, 0) and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$ -axis. If its minor axis subtends an angle $60 °$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________. [2023]

Q 18 :

Let $C$ be the largest circle centred at $(2, 0)$ and inscribed in the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If $(1, α)$ lies on $C$ , then $10 α^{2}$ is equal to _______ . [2023]

Q 19 :

Let a tangent to the curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _______ . [2023]

Q 20 :

Let the line $y - x = 1$ intersect the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is: [2026]

Q 22 :

An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

Q 23 :

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

Q 24 :

If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $a b + 18 r^{2}$ is. [2026]

Q 25 :

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Q 26 :

Let S and S′ be the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If ${(S P)}^{2} + {(S' P)}^{2} - S P \cdot S' P = 37,$ then $α^{2} + β^{2}$ is equal to: [2026]

Q 27 :

Let the foci of the Ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and Hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the Hyperbola is:

Q 30 :

If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to:

Q 32 :

The length of focal chord AB of ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ is $(Given A = (\frac{8}{5}, \frac{3 \sqrt{3}}{5}))$

Q 33 :

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a^{2} = f o g (10), b^{2} = g o f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then $8 e^{2} + l^{2}$ is equal to

Q 34 :

Let L be a common tangent line to the curves $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and $x^{2} + y^{2} = \frac{31}{4}$ . Then, the square of the slope of the line L is ________

Want Us to Email you About Special Offers & Updates?

	Column-I		Column-II
(I)	If the coordinates of a point are $(4 \tan θ, 3 \sec θ)$ where $θ (θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$ is parameter then the points lies on a conic whose eccentricity is	P	$\sqrt{3}$
(II)	If an ellipse has the length of major axis 10 units and semi minor axis length 4 units, then this ellipse has eccentricity	Q	$\frac{4}{5}$
(III)	If AB is double ordinate of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{9} = 1$ such that triangle OAB is an equilateral triangle of side '2' then eccentricity of hyperbola is (where O is centre of Hyperbola)	R	$\frac{5}{3}$
(IV)	If the foci of ellipse $\frac{x^{2}}{K^{2} a^{2}} + \frac{y^{2}}{a^{2}} = 1$ and hyperbol $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$ coincide then K can be	S	$\frac{3}{5}$
		T	$\sqrt{\frac{13}{3}}$

Topic Question Set

Q 11 : Consider ellipses Ek:kx2+k2y2=1, k=1,2,…,20. Let Ck be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse Ek. If rk is the radius of the circle Ck, then the value of ∑k=1201rk2 is [2023]

Q 12 : If the radius of the largest circle with centre (2, 0) inscribed in the ellipse x2+4y2=36 is r, then 12r2 is equal to [2023]

Q 13 : Let P(237, 67), Q, R and S be four points on the ellipse 9x2+4y2=36. Let PQ and RS be mutually perpendicular and pass through the origin. If 1(PQ)2+1(RS)2=pq, where p and q are coprime, then p+q is equal to [2023]

(1)

Q 15 : If the maximum distance of normal to the ellipse x24+y2b2=1,b<2, from the origin is 1, then the eccentricity of the ellipse is: [2023]

Q 16 : Let an ellipse with centre (1, 0) and latus rectum of length 12 have its major axis along the x-axis. If its minor axis subtends an angle 60° at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________. [2023]

Q 17 : The line x=8 is the directrix of the ellipse E:x2a2+y2b2=1 with the corresponding focus (2,0). If the tangent to E at the point P in the first quadrant passes through the point (0,43) and intersects the x-axis at Q, then (3PQ)2 is equal to __________ . [2023]

Q 18 : Let C be the largest circle centred at (2,0) and inscribed in the ellipse x236+y216=1. If (1,α) lies on C, then 10α2 is equal to _______ . [2023]

Q 19 : Let a tangent to the curve 9x2+16y2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _______ . [2023]

Q 20 : Let the line y-x=1 intersect the ellipse x22+y21=1 at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is: [2026]

Q 21 : Let each of the two ellipses E1:x2a2+y2b2=1, (a>b) and E2:x2A2+y2B2=1, (A<B) have eccentricity 45. Let the lengths of the latus recta of E1 and E2 be l1 and l2, respectively, such that 2l1 2=9l2. If the distance between the foci of E1 is 8, then the distance between the foci of E2 is [2026]

Q 22 : An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

Q 23 : Let the length of the latus rectum of an ellipse x2a2+y2b2=1, (a>b), be 30. If its eccentricity is the maximum value of the function f(t)=-34+2t-t2, then (a2+b2) is equal to [2026]

Q 24 : If the points of intersection of the ellipses x2+2y2-6x-12y+23=0 and 4x2+2y2-20x-12y+35=0 lie on a circle of radius r and centre (a,b), then the value of ab+18r2 is. [2026]

Q 25 : If the line ax+4y=7, where a∈R touches the ellipse 3x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Q 26 : Let S and S′ be the foci of the ellipse x225+y29=1 and P(α,β) be a point on the ellipse in the first quadrant. If (SP)2+(S'P)2-SP·S'P=37, then α2+β2 is equal to: [2026]

Q 27 : Let the foci of the Ellipse x216+y27=1 and Hyperbola x2144-y2α=125 coincide. Then the length of the latus rectum of the Hyperbola is:

Q 29 : Let P(237,67), Q, R and S be four points on ellipse 9x2+4y2=36. Let PQ and RS be mutually perpendicular chords and pass through the centre of ellipse, then the value of [50PQ2+50RS2]=____ where [.] denotes GIF

Q 30 : If the length of the latus rectum of the ellipse x2+4y2+2x+8y-λ=0 is 4, and l is the length of its major axis, then λ+l is equal to:

Q 32 : The length of focal chord AB of ellipse x24+y23=1 is (Given A=(85,335))

Q 33 : Let f(x)=x2+9, g(x)=xx-9 and a2=fog(10), b2=gof(3). If e and l denote the eccentricity and the length of the latus rectum of the ellipse x2a2+y2b2=1, then 8e2+l2 is equal to

Q 34 : Let L be a common tangent line to the curves x29+y24=1 and x2+y2=314. Then, the square of the slope of the line L is ________

Want Us to Email you About Special Offers & Updates?

Q 12 :

If the radius of the largest circle with centre (2, 0) inscribed in the ellipse $x^{2} + 4 y^{2} = 36$ is $r$ , then $12 r^{2}$ is equal to [2023]

Q 15 :

If the maximum distance of normal to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1, b < 2,$ from the origin is 1, then the eccentricity of the ellipse is: [2023]

Q 16 :

Let an ellipse with centre (1, 0) and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$ -axis. If its minor axis subtends an angle $60 °$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________. [2023]

Q 18 :

Let $C$ be the largest circle centred at $(2, 0)$ and inscribed in the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 .$ If $(1, α)$ lies on $C$ , then $10 α^{2}$ is equal to _______ . [2023]

Q 19 :

Let a tangent to the curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _______ . [2023]

Q 20 :

Let the line $y - x = 1$ intersect the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is: [2026]

Q 22 :

An ellipse has its center at (1, −2), one focus at (3, −2) and one vertex at (5, −2). Then the length of its latus rectum is : [2026]

Q 23 :

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

Q 24 :

If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $a b + 18 r^{2}$ is. [2026]

Q 25 :

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Q 26 :

Let S and S′ be the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ and $P (α, β)$ be a point on the ellipse in the first quadrant. If ${(S P)}^{2} + {(S' P)}^{2} - S P \cdot S' P = 37,$ then $α^{2} + β^{2}$ is equal to: [2026]

Q 27 :

Let the foci of the Ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and Hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the Hyperbola is:

Q 30 :

If the length of the latus rectum of the ellipse $x^{2} + 4 y^{2} + 2 x + 8 y - λ = 0$ is 4, and $l$ is the length of its major axis, then $λ + l$ is equal to:

Q 32 :

The length of focal chord AB of ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ is $(Given A = (\frac{8}{5}, \frac{3 \sqrt{3}}{5}))$

Q 33 :

Let $f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}$ and $a^{2} = f o g (10), b^{2} = g o f (3)$ . If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then $8 e^{2} + l^{2}$ is equal to

Q 34 :

Let L be a common tangent line to the curves $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and $x^{2} + y^{2} = \frac{31}{4}$ . Then, the square of the slope of the line L is ________