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

Q 31 :

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

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(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 32 :

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

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(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 33 :

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

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(1)

Q 34 :

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

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(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 35 :

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]

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

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(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 36 :

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]

0%

(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 37 :

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]

0%

(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 38 :

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]

0%

(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 39 :

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]

0%

(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 40 :

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]

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

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(2)

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

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

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

$m_{O B} = \tan θ = \frac{1}{4}$

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

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