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

Q 21 :

Let the product of the focal distances of the point $(\sqrt{3}, \frac{1}{2})$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b), be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipse is [2025]

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

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(1)

Product of focal distances from point $(\sqrt{3}, \frac{1}{2})$

$= (a + e x_{1}) (a - e x_{1})$

$\Rightarrow a^{2} - 3 e^{2} = \frac{7}{4}$ [ $∵ x_{1} = \sqrt{3}$ ]

$\Rightarrow 4 a^{2} = 7 + 12 e^{2}$

Also, $(\sqrt{3}, \frac{1}{2})$ lies on ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{3}{a^{2}} + \frac{1}{4 b^{2}} = 1 \Rightarrow \frac{3}{a^{2}} + \frac{1}{4 a^{2} (1 - e^{2})} = 1$ [ $∵ b^{2} = a^{2} (1 - e^{2})$ ]

$\Rightarrow 12 (1 - e^{2}) + 1 = 4 a^{2} (1 - e^{2})$

$\Rightarrow 13 - 12 e^{2} = (7 + 12 e^{2}) (1 - e^{2})$

$\Rightarrow 13 - 12 e^{2} = 7 + 12 e^{2} - 7 e^{2} - 12 e^{4}$

$\Rightarrow 12 e^{4} - 17 e^{2} + 6 = 0$

$\Rightarrow e^{2} = \frac{3}{4} or \frac{2}{3} \Rightarrow e = \frac{\sqrt{3}}{2} or \sqrt{\frac{2}{3}}$

$∴$ Required difference = $| \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} |$ $= \frac{3 - 2 \sqrt{2}}{2 \sqrt{3}}$

Q 22 :

The equation of the chord, of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid-point is (3, 1) is : [2025]

4x + 122y = 134
5x + 16y = 31
25x + 101y = 176
48x + 25y = 169

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(4)

Given: Ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and mid-point is (3, 1).

The equation of chord with given middle point is given by

T = $S_{1}$

$\Rightarrow \frac{3 x}{25} + \frac{1 y}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$

$\Rightarrow \frac{3 x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$

$\Rightarrow 48 x + 25 y = 144 + 25$

$\Rightarrow 48 x + 25 y = 169$ .

Q 23 :

Let the ellipse $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $E_{2} : \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$ , A < B have same eccentricity $\frac{1}{\sqrt{3}}$ . Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ , and the distance between the foci of $E_{1}$ be 4. If $E_{1}$ and $E_{2}$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals : [2025]

$6 \sqrt{6}$
$\frac{12 \sqrt{6}}{5}$
$\frac{24 \sqrt{6}}{5}$
$\frac{18 \sqrt{6}}{5}$

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(3)

We have, $2 a e = 4 \Rightarrow 2 a (\frac{1}{\sqrt{3}}) = 4 \Rightarrow a = 2 \sqrt{3}$

$b^{2} = a^{2} (1 - e^{2}) = 12 (1 - \frac{1}{3}) = 8$

Now, $\Rightarrow \frac{2 b^{2}}{a} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}}$

$\Rightarrow \frac{2 \cdot 8}{2 \sqrt{3}} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}} \Rightarrow A^{2} = 2 B$

Alao, $\Rightarrow A^{2} = B^{2} (1 - e^{2}) \Rightarrow 2 B = B^{2} \cdot \frac{2}{3} \Rightarrow B = 3 \Rightarrow A^{2} = 6$

$∴ E_{1} : \frac{x^{2}}{12} + \frac{y^{2}}{8} = 1$ ... (i)

and $E_{2} : \frac{x^{2}}{6} + \frac{y^{2}}{9} = 1$ ... (ii)

Solving (i) and (ii), we get

$A (\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), B (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), C (\frac{\sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}}), D (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}})$

which form a rectangle.

$∴$ Required area $\frac{12}{\sqrt{5}} \times \frac{2 \sqrt{6}}{\sqrt{5}} = \frac{24 \sqrt{6}}{5}$

Q 24 :

If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

22
26
20
18

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(1)

Let AB is a chord and M is the mid-point.

If $M (\sqrt{2}, \frac{4}{3})$ then equation of AB is

$T = S_{1} \Rightarrow \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} - 1 = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1$

$\Rightarrow \frac{x \sqrt{2}}{9} + \frac{y}{4} (\frac{4}{3}) = \frac{{(\sqrt{2})}^{2}}{9} + \frac{{(\frac{4}{3})}^{2}}{4}$

$\Rightarrow \frac{\sqrt{2} x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9}$

$\Rightarrow \sqrt{2} x + 3 y = 6 \Rightarrow y = \frac{6 - \sqrt{2} x}{3}$

Putting in ellipse, we get $\frac{x^{2}}{9} + \frac{{(6 - \sqrt{2} x)}^{2}}{9 \times 4} = 1$

$\Rightarrow 4 x^{2} + 36 + 2 x^{2} - 12 \sqrt{2} x = 36$

$\Rightarrow 6 x^{2} - 12 \sqrt{2} x = 0 \Rightarrow 6 x (x - 2 \sqrt{2}) = 0$

$\Rightarrow x = 0 and x = 2 \sqrt{2}$

So, y = 2 and $y = \frac{2}{3}$

$∴$ Length of the chord = $= \sqrt{{(2 \sqrt{2} - 0)}^{2} + {(\frac{2}{3} - 2)}^{2}}$

$= \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2}{3} \sqrt{22}$

So, $α = 22$ .

Q 25 :

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to : [2025]

46
58
37
72

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(2)

We have, equation of ellipse as, $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

Equation of chord with mid-point $(\frac{5}{2}, \frac{1}{2})$ is

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

$\Rightarrow \frac{5}{2} (\frac{x}{9}) + \frac{1}{2} (\frac{y}{4}) = \frac{25}{36} + \frac{1}{16}$

$\Rightarrow \frac{5 x}{18} + \frac{y}{8} = \frac{109}{144}$

$\Rightarrow$ 40x + 18y = 109

On comparing with given equation $α x + β y = 109$ , we get

$α = 40 and β = 18$

$∴ α + β = 58$ .

Q 26 :

Let C be the circle $x^{2} + {(y - 1)}^{2} = 2$ , $E_{1}$ and $E_{2}$ be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line x + y = 3 touch the curves C, $E_{1}$ and $E_{2}$ at $P (x_{1}, y_{1})$ , $Q (x_{2}, y_{2})$ and $R (x_{3}, y_{3})$ respectively. Given that P is the mid-point of the line segment QR and $P Q = \frac{2 \sqrt{2}}{3}$ , the value of $9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})$ is equal to __________. [2025]

0%

(46)

(a) Solving the line x + y = 3, and the circle $x^{2} + {(y - 1)}^{2} = 2$

Substitute y = 3 – x

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

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

$\Rightarrow x = 1 \Rightarrow y = 2$

So, $P = (x_{1}, y_{1}) = (1, 2) \Rightarrow x_{1} y_{1} = 2$

Use mid-point condition

Let $Q = (x_{2}, y_{2})$ , $R = (x_{3}, y_{3})$ .

Since P is the mid-point of QR

$\Rightarrow x_{2} + x_{3} = 2 x_{1} = 2, y_{2} + y_{3} = 2 y_{1} = 4$

So, we can write : $x_{3} = 2 - x_{2}, y_{3} = 4 - y_{2}$

(b) Given

$P Q = \frac{2 \sqrt{2}}{3} \Rightarrow P Q^{2} = {(x_{2} - 1)}^{2} + {(y_{2} - 2)}^{2} = \frac{8}{9}$

Let's denote : $x_{2} = a, y_{2} = b, x_{3} = 2 - a, y_{3} = 5 - b$

${(a - 1)}^{2} + {(b - 2)}^{2} = \frac{8}{9}$

$\Rightarrow a^{2} - 2 a + 1 + b^{2} - 4 b + 4 = \frac{8}{9}$

$\Rightarrow a^{2} + b^{2} - 2 a - 4 b + 5 = \frac{8}{9}$

$\Rightarrow 9 a^{2} + 9 b^{2} - 18 a - 36 b + 37 = 0$

Hence, $a = \frac{5}{3}, b = \frac{4}{3}$

$x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} = 2 + a b + (2 - a) (4 - b)$

$9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3}) = 9 (10 + 2 a b - 2 b - 4 a)$

= 90 + 18ab – 18b – 36a = 46.

Q 27 :

Let $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ be an ellipse. Ellipse $E_{i}' s$ are constructed such that their centres and eccentricities are same as that of $E_{1}$ , and the length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1} (i \geq 1)$ . If $A_{i}$ is the area of the ellipse $E_{i}$ , then $\frac{5}{π} (\sum_{i = 1}^{\infty} A_{i})$ , is equal to __________. [2025]

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(54)

Given, $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{\sqrt{5}}{3}$

As length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1}$ .

$E_{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{a^{2}}{4^{2}}} = \frac{\sqrt{5}}{3}$ [ $∵$ Eccentricities are same]

$\Rightarrow a = \frac{4}{3}$ $∴ E_{2} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{4} = 1$

Now, $E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{b^{2}}{\frac{16}{9}}} = \frac{\sqrt{5}}{3}$

$\Rightarrow b^{2} = \frac{64}{81}$ $∴ E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{\frac{64}{81}} = 1$

Now, $A_{1} = π a b = π \times 3 \times 2 = 6 π$ ,

$A_{2} = π \times \frac{4}{3} \times 2 = \frac{8}{3} π$

and $A_{3} = π \times \frac{4}{3} \times \frac{8}{9} = \frac{32}{27} π$

$∴ \sum_{i = 1}^{\infty} A_{i} = 6 π + \frac{8}{3} π + \frac{32}{27} π + . . . \infty$

$= \frac{6 π}{1 - \frac{4}{9}} = \frac{54 π}{5}$ $[∵ S_{\infty} = \frac{a}{1 - r}]$

So, $\frac{5}{π} \sum_{i = 1}^{\infty} A_{i} = \frac{5}{π} \times \frac{54 π}{5} = 54$ .

Q 28 :

In a group of 100 persons, 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons who speak only English is $α$ and the number of persons who speak only Hindi is $β$ , then the eccentricity of the ellipse $25 (β^{2} x^{2} + α^{2} y^{2}) = α^{2} β^{2}$ is [2023]

$\frac{\sqrt{117}}{12}$
$\frac{\sqrt{119}}{12}$
$\frac{\sqrt{129}}{12}$
$\frac{3 \sqrt{15}}{12}$

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(2)

Let $γ$ be the number of persons who speak both English and Hindi.

According to the question, we have

$α + β + γ = 100 (i)$

$α + γ = 75 (ii)$

and $β + γ = 40 (iii)$

Solving (i), (ii) and (iii), we get

$α = 60,$ $β = 25$ and $γ = 15$

So, equation of the given ellipse becomes,

$25 (25^{2} x^{2} + 60^{2} y^{2}) = 60^{2} \times 25^{2}$

$\Rightarrow \frac{x^{2}}{{(60 / 5)}^{2}} + \frac{y^{2}}{{(25 / 5)}^{2}} = 1 \Rightarrow \frac{x^{2}}{12^{2}} + \frac{y^{2}}{5^{2}} = 1$

So, eccentricity of ellipse $= \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{5^{2}}{12^{2}}} = \frac{\sqrt{119}}{12}$

Q 29 :

Let the ellipse $E : x^{2} + 9 y^{2} = 9$ intersect the positive $x$ - and $y$ -axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $\frac{m}{n}$ , where m and n are coprime, then $m - n$ is equal to [2023]

17
15
18
16

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(1)

We have, $E : \frac{x^{2}}{9} + \frac{y^{2}}{1} = 1$

$A \equiv (3, 0)$

$B \equiv (0, 1)$

Equation of line passing through A(3, 0) and B(0, 1) is $\frac{y - 0}{x - 3} = \frac{1 - 0}{0 - 3} \Rightarrow x + 3 y = 3$ ...(i)

The equation of the circle with radius 3 is $x^{2} + y^{2} = 9$ ...(ii)

From (i) and (ii), we get
${(3 - 3 y)}^{2} + y^{2} = 9 \Rightarrow 10 y^{2} - 18 y = 0$ $\Rightarrow y = 0, \frac{9}{5}$

$∴$ $Area of triangle O P A = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10} = \frac{m}{n}$

$∴$ $m - n = 17$

Q 30 :

Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285 .$ Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]

$\frac{π}{12}$
$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{3}$

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(4)

$Given ellipse is \frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$

$Equation of tangent to the ellipse is given by$

$y = m x \pm \sqrt{19 m^{2} + 15}$ ...(i)

$Equation of tangent to the circle x^{2} + y^{2} = 16 is given by$

$y = m x \pm 4 \sqrt{1 + m^{2}}$ ...(ii)

$From (i) and (ii), m x \pm 4 \sqrt{1 + m^{2}} = m x \pm \sqrt{19 m^{2} + 15}$

$\Rightarrow 4 \sqrt{1 + m^{2}} = \sqrt{19 m^{2} + 15}$ $\Rightarrow 16 (1 + m^{2}) = 19 m^{2} + 15$

$\Rightarrow 16 + 16 m^{2} = 19 m^{2} + 15$

$\Rightarrow 1 = 3 m^{2} \Rightarrow m^{2} = \frac{1}{3} \Rightarrow m = \pm \frac{1}{\sqrt{3}}$

$\Rightarrow \tan θ = \pm \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6} with x-axis$

$∴$ $Common tangent inclined to the minor axis of the ellipse at the angle \frac{π}{3} .$

Topic Question Set

Q 21 :

Let the product of the focal distances of the point $(\sqrt{3}, \frac{1}{2})$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b), be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipse is [2025]

Q 22 :

The equation of the chord, of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid-point is (3, 1) is : [2025]

Q 24 :

If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

Q 25 :

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to : [2025]

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Q 30 :

Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285 .$ Then the common tangents are inclined to the minor axis of the ellipse at the angle [2023]

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