Let the ellipse, E1:x2a2+y2b2=1, a > b and E2:x2A2+y2B2=1, A < B have same eccentricity 13. Let the product of their lengths of latus rectums be 323, and the distance between the foci of E1 be 4. If E1 and E2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals:

2ae = 42a13=4⇒a=23⇒1−b212=13⇒b2=8 Now 2b2a⋅2A2B=323⇒28232A2B=323⇒A2=2B1−A2B2=13⇒1−2BB2=13⇒B=3⇒A2=6x212+y28=1…..(1)x26+y29=1…..(2)On solving (1) & (2) we get(x,y)≡65,65,−65,65,65,−65,−65,−65The four points are vertices of rectangle and its area = 2465

Let the ellipse, E1:x2a2+y2b2=1, a > b and E2:x2A2+y2B2=1, A < B have same eccentricity 13. Let the product of their lengths of latus rectums be 323, and the distance between the foci of E1 be 4. If E1 and E2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals:

2ae = 42a13=4⇒a=23⇒1−b212=13⇒b2=8 Now 2b2a⋅2A2B=323⇒28232A2B=323⇒A2=2B1−A2B2=13⇒1−2BB2=13⇒B=3⇒A2=6x212+y28=1…..(1)x26+y29=1…..(2)On solving (1) & (2) we get(x,y)≡65,65,−65,65,65,−65,−65,−65The four points are vertices of rectangle and its area = 2465

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 E₁ be 4. If E₁ and E₂ meet at A, B, C and D, then the area of the quadrilateral ABCD equals:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

6 $\sqrt{6}$

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

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

2ae = 4
$\begin{array}{l} 2 a (\frac{1}{\sqrt{3}}) = 4 \\ \Rightarrow a = 2 \sqrt{3} \\ \Rightarrow 1 - \frac{b^{2}}{12} = \frac{1}{3} \Rightarrow b^{2} = 8 \\ Now \frac{2 b^{2}}{a} \cdot \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}} \Rightarrow 2 (\frac{8}{2 \sqrt{3}}) \frac{2 A^{2}}{B} = \frac{32}{\sqrt{3}} \\ \Rightarrow A^{2} = 2 B \\ 1 - \frac{A^{2}}{B^{2}} = \frac{1}{3} \Rightarrow 1 - \frac{2 B}{B^{2}} = \frac{1}{3} \Rightarrow B = 3 \\ \Rightarrow A^{2} = 6 \\ \frac{x^{2}}{12} + \frac{y^{2}}{8} = 1 \dots . . (1) \\ \frac{x^{2}}{6} + \frac{y^{2}}{9} = 1 \dots . . (2) \end{array}$
On solving (1) & (2) we get
$(x, y) \equiv (\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}), (\frac{\sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}}), (\frac{- \sqrt{6}}{\sqrt{5}}, \frac{- 6}{\sqrt{5}})$
The four points are vertices of rectangle and its area = $\frac{24 \sqrt{6}}{5}$