JEE Main 2025 (Online) 29th January Morning ShiftQuestion 1
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 $\frac1{\sqrt3}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt3}$, 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 :
Ellipse
Solution
Concept. Both ellipses are symmetric about both axes, so the four common points form a rectangle $(\pm x,\pm y)$. Formulas used. (1) $a>b$: foci distance $2ae$, $b^2=a^2(1-e^2)$, latus rectum $\frac{2b^2}{a}$(2) $A<B$: $A^2=B^2(1-e^2)$, latus rectum $\frac{2A^2}{B}$Step 1 - $E_1$, by (1). $2a\cdot\frac1{\sqrt3}=4\Rightarrow a=2\sqrt3$, $a^2=12$, $b^2=12\cdot\frac23=8$. Latus rectum $\frac{16}{2\sqrt3}=\frac8{\sqrt3}$. Step 2 - $E_2$, by (2). Its latus rectum $=\frac{32/\sqrt3}{8/\sqrt3}=4$. $A^2=\frac23B^2$, so $\frac{2A^2}{B}=\frac43B=4\Rightarrow B=3$, $A^2=6$. Step 3 - Intersection. Let $u=x^2$, $v=y^2$: $$\tfrac{u}{12}+\tfrac v8=1\Rightarrow2u+3v=24;\qquad\tfrac u6+\tfrac v9=1\Rightarrow3u+2v=18.$$ Solving: $v=\frac{36}{5}$, $u=\frac65$.Step 4 - Area. Rectangle of sides $2|x|$ and $2|y|$: $$4\sqrt{\tfrac65\cdot\tfrac{36}{5}}=4\cdot\tfrac{6\sqrt6}{5}=\tfrac{24\sqrt6}{5}.$$ Answer: D