Let the ellipse E_1: x^2 a^2+ y^2 ~b^2=1, agt b and E_2: x^2 ~A^2+ y^2 ~B^2=1, ~A lt B have same eccentricity 1 √3. Let the product of their lengths…
JEE Main 2025 — Mathematics Coordinate Geometry
2025mcqhard
Let the ellipse E1:a2x2+b2y2=1,a>b and E2:A2x2+B2y2=1,A<B have same eccentricity 31. Let the product of their lengths of latus rectums be 332, 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 :
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
A
5126
B
66
C
5186
D
5246
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Solution
2ae=4⇒a=23⇒1−12b2=31⇒b2=8a2b2×B2A2=332⇒232×8×B2A2=332⇒BA2=2⇒A2=2B1−BA2=31⇒B=3⇒A2=6E1:12x2+8y2=1 ....(i) E1:6x2+9y2=1...(ii) On solving (i) & (ii) (x,y)=(56,56),(5−6,56),(56,5−6),(5−6,5−6) Four points are vertices of rectangle area =5246
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