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

Q 11 :
Let the latus rectum of the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{b^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + \sqrt{n})$ , where $l$ and m are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to __________. [2024]

0%

(182)

In right triangle OFP, we have $\tan 30 ° = \frac{P F}{O F}$

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

$e = \frac{\sqrt{3} b^{2}}{9}$ (Given $a^{2} = 9$ )

Squaring both sides, we get

$e^{2} = \frac{3 b^{4}}{81}$ ... (i)

and $e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow e^{2} = 1 + \frac{b^{2}}{9}$ ... (ii)

Comparing equation (i) and (ii), we get

$\frac{3 b^{4}}{81} = 1 + \frac{b^{2}}{9} \Rightarrow b^{4} = 27 + 3 b^{2}$

$\Rightarrow b^{4} - 3 b^{2} - 27 = 0 \Rightarrow b^{2} = \frac{3}{2} (1 + \sqrt{13})$

According to given condition, $b^{2} = \frac{l}{m} (1 + \sqrt{n})$

On comparing, we get $l$ = 3, m = 2 and n = 13

$\Rightarrow l^{2} + m^{2} + n^{2} = {(3)}^{2} + {(2)}^{2} + {(13)}^{2} = 9 + 4 + 160 = 182$ .

Q 12 :
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 be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

0%

(51)

Here ae = 5 and $\frac{2 b^{2}}{a} = \sqrt{50}$

$\frac{b^{2}}{a} = \frac{5 \sqrt{2}}{2}$

$a^{2} e^{2} = a^{2} - b^{2} \Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}}$

$25 = a^{2} - \frac{5 \sqrt{2}}{2} a \Rightarrow 2 a^{2} - 5 \sqrt{2} a - 50 = 0$

$\Rightarrow [a - 5 \sqrt{2}] [2 a + 5 \sqrt{2}] = 0 \Rightarrow a = 5 \sqrt{2} or a = \frac{- 5 \sqrt{2}}{2}$

$\Rightarrow e = \frac{5}{a} = \frac{1}{\sqrt{2}}$

Also, $b^{2} = 5 \sqrt{2} \times \frac{5 \sqrt{2}}{2} = 25 \Rightarrow b = 5$

Now, $e = \sqrt{1 + \frac{a^{2} b^{2}}{b^{2}}} = \sqrt{1 + a^{2}} \Rightarrow e^{2} = 1 + a^{2} = 51$ .

Q 13 :
Let one focus of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$ . If e and $l$ respectively are the eccentricity and the length of the latus rectum of H, then $9 (e^{2} + l)$ is equal to : [2025]

16
15
12
14

1

2

3

4

(1)

We have, $a e = \sqrt{10} and \frac{a}{e} = \frac{9}{\sqrt{10}}$

$\Rightarrow a^{2} = 9 and e = \frac{\sqrt{10}}{3}$

Now, ${(a e)}^{2} = a^{2} + b^{2} \Rightarrow 10 = 9 + b^{2} \Rightarrow b^{2} = 1$

and $l = \frac{2 b^{2}}{a} = \frac{2 (1)}{3} = \frac{2}{3}$

$∴ 9 (e^{2} + l) = 9 (\frac{10}{9} + \frac{2}{3}) = 10 + 6 = 16$

Q 14 :
Let the sum of the focal distances of the point P(4, 3) on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be $8 \sqrt{\frac{5}{3}}$ . If for H, the length of the latus rectum is $l$ and the product of the focal distances of the point P is m, then $9 l^{2} + 6 m$ is equal to : [2025]

187
184
185
186

1

2

3

4

(3)

Given, hyperbola $(H) : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Sum of focal distance of P(4, 3) = $8 \sqrt{\frac{5}{3}}$

$\Rightarrow 2 e x = 8 \sqrt{\frac{5}{3}} \Rightarrow 2 e (4) = 8 \sqrt{\frac{5}{3}} \Rightarrow e^{2} = \frac{5}{3}$

$∵ b^{2} = a^{2} (\frac{5}{3} - 1) \Rightarrow b^{2} = \frac{2}{3} a^{2}$ ... (i)

$\Rightarrow \frac{16}{a^{2}} - \frac{9}{b^{2}} = 1$ ... (ii)

Using (i) and (ii), we get

$\Rightarrow a^{2} = \frac{5}{2}, b^{2} = \frac{5}{3}$

Now, length of latus rectum, $l = \frac{2 b^{2}}{a} \Rightarrow l^{2} = \frac{4 b^{4}}{a^{2}}$

$\Rightarrow l^{2} = \frac{4 \times {(5 / 3)}^{2}}{5 / 2} \Rightarrow 9 l^{2} = 40$

Also, $m = (e x + a) (e x - a) = {(e x)}^{2} - {(a)}^{2}$

$= \frac{5}{3} (16) - \frac{5}{2} = \frac{145}{6}$

$\Rightarrow 6 m = 145$

$∴ 9 l^{2} + 6 m = 40 + 145 = 185$ .

Q 15 :
Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{25} = 1$ and the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1$ , respectively. If b < 5 and $e_{1} e_{2} = 1$ , then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is : [2025]

$\frac{\sqrt{3}}{2}$
$\frac{4}{5}$
$\frac{3}{5}$
$\frac{\sqrt{7}}{4}$

1

2

3

4

(3)

We have, $e_{1}^{2} = 1 - \frac{b^{2}}{25} and e_{2}^{2} = 1 + \frac{b^{2}}{16}$

$∵ e_{1} e_{2} = 1$ (Given)

$∴ e_{1}^{2} e_{2}^{2} = 1$

$\Rightarrow (1 - \frac{b^{2}}{25}) (1 + \frac{b^{2}}{16}) = 1$

$\Rightarrow 1 + \frac{b^{2}}{16} - \frac{b^{2}}{25} - \frac{b^{4}}{400} = 1$

$\Rightarrow \frac{9 b^{2}}{400} = \frac{b^{4}}{400}$

$\Rightarrow b^{2} = 9$

We get, $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1 and \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

Here, $e_{1} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} and e_{2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$

Foci : $(0, \pm 4) and (\pm 5, 0)$

The ellipse passing through all four foci is $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$

Now, the eccentricity is given by $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ .

Q 16 :
Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

$\frac{25}{6}$
$\frac{24}{5}$
$\frac{144}{5}$
$\frac{288}{5}$

1

2

3

4

(4)

Distance between foci $\sqrt{{(1 - 1)}^{2} + {(14 + 12)}^{2}}$

$\Rightarrow 2 b e = 26 \Rightarrow b e = 13$

Mid-point of foci $= (\frac{1 + 1}{2}, \frac{14 - 12}{2}) = (1, 1)$

Equation of hyperbola is $\frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y - 1)}^{2}}{b^{2}} = 1$

Hyperbola passes through (1, 6)

$∴ \frac{{(6 - 1)}^{2}}{b^{2}} = 1 \Rightarrow b^{2} = 25$

Now, $a^{2} = b^{2} (e^{2} - 1) = b^{2} e^{2} - b^{2} = 169 - 25 = 144$

$∴ a^{2} = 144 and b^{2} = 25$

$∴$ Length of latus rectum $= \frac{2 a^{2}}{b} = \frac{2 \times 144}{5} = \frac{288}{5}$

Q 17 :
Let $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ . Let the distane between the foci of E and the foci of H be $2 \sqrt{3}$ . If a – A = 2, and the ratio of the eccentricities of E and H is $\frac{1}{3}$ , then the sum of the lengths of their latus rectums is equal to: [2025]

9
7
10
8

1

2

3

4

(4)

We have, $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , whose foci are ( $\pm$ ae, 0) and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ , whose foci are ( $\pm$ Ae', 0).

Given, $2 a e = 2 \sqrt{3} and 2 A e' = 2 \sqrt{3}$

$\Rightarrow a e = \sqrt{3} and A e' = \sqrt{3}$

$\Rightarrow \frac{a e}{A e'} = 1 \Rightarrow \frac{e}{e'} = \frac{A}{a}$

$\Rightarrow \frac{1}{3} = \frac{A}{a} \Rightarrow a = 3 A$ $[Given \frac{e}{e'} = \frac{1}{3}]$

Now, a – A = 2 $\Rightarrow$ 2A = 1 [a = 3A]

$\Rightarrow A = 1, a = 3$

$∴ e = \frac{1}{\sqrt{3}} and e' = \sqrt{3}$

Also, $b^{2} = a^{2} (1 - e^{2}) and B^{2} = A^{2} ({(e')}^{2} - 1)$

$\Rightarrow b^{2} = 9 (1 - {(\frac{1}{\sqrt{3}})}^{2}) and B^{2} = {(1)}^{2} ({(\sqrt{3})}^{2} - 1)$

$\Rightarrow b^{2} = 6 and B^{2} = 2$

$∴$ Sum of latus rectum $= \frac{2 b^{2}}{a} + \frac{2 B^{2}}{A} = 8$ .

Q 18 :
If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

x + 9y = 36
9x – 9y = 32
4x – 9y = 12
6x – 9y = 20

1

2

3

4

(4)

We have, $x^{2} + y^{2} - 8 x = 0$ ... (i)

and $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 \Rightarrow 4 x^{2} - 9 y^{2} = 36$ ... (ii)

Solving (i) and (ii), we get

$4 x^{2} - 9 (8 x - x^{2}) = 36$

$\Rightarrow 13 x^{2} - 72 x - 36 = 0$

$\Rightarrow (13 x + 6) (x - 6) = 0$

$\Rightarrow x = \frac{- 6}{13}, 6$

$∴ x = 6$ $[∵ x = \frac{- 6}{13} not possible]$

From (i), $y^{2} = 8 (6) - {(6)}^{2} = 48 - 36 = 12 \Rightarrow y = \pm \sqrt{12}$

$∴$ $(6, \sqrt{12})$ and $(6, - \sqrt{12})$ are the points of intersection of circle and hyperbola.

Let $P (α, β)$ be the point moves on the line 2x – 3y + 4 = 0 such that

$2 α - 3 β + 4 = 0$ ... (iii)

Centroid of $△$ PAB is given by $(\frac{α + 6 + 6}{3}, \frac{β}{3}) = (h, k)$

$∴ α = 3 h - 12 and β = 3 k$

From (iii), 2(3h –12) – 3(3k) + 4 = 0

$\Rightarrow 6 h - 24 - 9 k + 4 = 0 \Rightarrow 6 h - 9 k = 20$

i.e., 6x – 9y = 20.

Q 19 :
Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

0%

(120)

We have, $P (4, 2 \sqrt{3})$

Now, $P S_{1} \cdot P S_{2} = 32$ ... (i)

where $S_{1} = (a e, 0) and S_{2} = (- a e, 0)$

ALso, $| P S_{1} - P S_{2} | = 2 a$

Now, $P (4, 2 \sqrt{3})$ lies on H

$∴ \frac{16}{a^{2}} - \frac{12}{b^{2}} = 1$

$\Rightarrow 16 b^{2} - 12 a^{2} = a^{2} b^{2}$ ... (ii)

Now, $| P S_{1} - P S_{2} |^{2} = 4 a^{2}$

$\Rightarrow P S_{1}^{2} + P S_{2}^{2} - 2 P S_{1} \cdot P S_{2} = 4 a^{2}$

$\Rightarrow {(a e - 4)}^{2} + 12 + {(a e + 4)}^{2} + 12 - 64 = 4 a^{2}$

$\Rightarrow 2 a^{2} e^{2} - 8 = 4 a^{2} \Rightarrow a^{2} + b^{2} - 4 = 2 a^{2}$

$\Rightarrow b^{2} - a^{2} = 4$ ... (iii)

From equation (ii) and (iii)

$\Rightarrow 16 (a^{2} + 4) - 12 a^{2} = a^{2} (a^{2} + 4)$

$\Rightarrow 16 a^{2} + 64 - 12 a^{2} = a^{4} + 4 a^{2}$

$\Rightarrow a^{4} = 64 \Rightarrow a^{2} = 8 ∴ b^{2} = 12$

Now, $p^{2} + q^{2} = 4 b^{2} + \frac{4 b^{4}}{a^{2}} = 120$ .

Q 20 :
If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]

0%

(141)

Equation of hyperbola is $\frac{{(x - 6)}^{2}}{a^{2}} - \frac{{(y - 2)}^{2}}{4 - a^{2}} = 1$

$\Rightarrow (4 - a^{2}) {(x - 6)}^{2} - a^{2} {(y - 2)}^{2} = a^{2} (4 - a^{2})$

$\Rightarrow (4 - a^{2}) x^{2} - a^{2} y^{2} + (12 a^{2} - 48) x + 4 a^{2} y + 144 - 44 a^{2} + a^{4} = 0$

On comparing with $3 x^{2} - y^{2} - α x + β y + γ = 0$ , we get $a^{2} = 1$ and $α = 36$ , $β = 4$ and $γ = 101$

$∴ α + β + γ = 141$

Topic Question Set

0%

Q 12 :
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 be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

0%

Q 16 :
Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 18 :
If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

Q 19 :
Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

0%

Q 20 :
If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : Let the latus rectum of the hyperbola x29–y2b2=1 subtend an angle of π3 at the centre of the hyperbola. If b2 is equal to lm(1+n), where l and m are co-prime numbers, then l2+m2+n2 is equal to __________. [2024]

0%

Q 12 : Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1, a > b be (±5, 0) and 50, respectively. Then, the square of the eccentricity of the hyperbola x2b2–y2a2b2=1 equals [2024]

0%

Q 13 : Let one focus of the hyperbola H : x2a2–y2b2=1 be at (10,0) and the corresponding directrix be x=910. If e and l respectively are the eccentricity and the length of the latus rectum of H, then 9(e2+l) is equal to : [2025]

Q 14 : Let the sum of the focal distances of the point P(4, 3) on the hyperbola H:x2a2–y2b2=1 be 853. If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then 9l2+6m is equal to : [2025]

Q 16 : Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 17 : Let E : x2a2+y2b2=1, a > b and H : x2A2–y2B2=1. Let the distane between the foci of E and the foci of H be 23. If a – A = 2, and the ratio of the eccentricities of E and H is 13, then the sum of the lengths of their latus rectums is equal to: [2025]

Q 18 : If A and B are the points of intersection of the circle x2+y2–8x=0 and the hyperbola x29–y24=1 and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of △PAB lies on the line : [2025]

Q 19 : Let the product of focal distances of the point P(4,23) on the hyperbola H : x2a2–y2b2=1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2+q2 is equal to __________. [2025]

0%

Q 20 : If the equation of the hyperbola with foci (4, 2) and (8, 2) is 3x2–y2–αx+βy+γ=0, then α+β+γ is equal to __________. [2025]

0%

Want Us to Email you About Special Offers & Updates?

Q 12 :
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 be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

Q 16 :
Let the foci of a hyperbola be (1, 14) and (1, –12). If it passes through the point (1, 6) then the length of its latus-rectum is : [2025]

Q 18 :
If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

Q 19 :
Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

Q 20 :
If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to __________. [2025]