Let the ellipse E_1: \fracx^2a^2 + \fracy^2b^2 = 1 , a \gt b and E_2: \fracx^2A^2 + \fracy^2B^2 = 1 , A \lt B have same eccentricity \frac1\sqrt3 . Let the product of their lengths of latus rectums be \frac32\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 PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q30 with Solution

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Correct answer: A

Write eccentricity relations for both ellipses

For $E_1: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$ , its eccentricity is $e_1=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{3}}.$ So,

\implies \frac{b^2}{a^2}=\frac{2}{3} \implies b^2=\frac{2a^2}{3}. $$ For $E_2: \dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ with $A<B$, the major axis is along $y$. Hence its eccentricity is $$e_2=\sqrt{1-\frac{A^2}{B^2}}=\frac{1}{\sqrt{3}}.$$ Thus, $$1-\frac{A^2}{B^2}=\frac{1}{3} \implies \frac{A^2}{B^2}=\frac{2}{3} \implies A^2=\frac{2B^2}{3}. $$ --- 2. **Use distance between foci of $E_1$** For $E_1$, focal distance is $2c=4$, so $$c=2.$$ But for an ellipse, $$c=ae= a\cdot \frac{1}{\sqrt{3}}.$$ Therefore, $$\frac{a}{\sqrt{3}}=2 \implies a=2\sqrt{3}. $$ Hence, $$a^2=12, \qquad b^2=\frac{2}{3}a^2=8. $$ So $$E_1:\frac{x^2}{12}+\frac{y^2}{8}=1.$$ --- 3. **Use product of latus recta** Length of latus rectum of $E_1$ is $$L_1=\frac{2b^2}{a}=\frac{2\cdot 8}{2\sqrt3}=\frac{8}{\sqrt3}. $$ Given $$L_1L_2=\frac{32}{\sqrt3},$$ so $$\frac{8}{\sqrt3}L_2=\frac{32}{\sqrt3} \implies L_2=4.$$ For $E_2$ (major axis along $y$), latus rectum length is $$L_2=\frac{2A^2}{B}=4.$$ Using $A^2=\dfrac{2B^2}{3}$, $$\frac{2\left(\frac{2B^2}{3}\right)}{B}=4 \implies \frac{4B}{3}=4 \implies B=3. $$ Then $$A^2=\frac{2B^2}{3}=\frac{2\cdot 9}{3}=6. $$ So $$E_2:\frac{x^2}{6}+\frac{y^2}{9}=1.$$ --- 4. **Find intersection points of the two ellipses** We solve $$\frac{x^2}{12}+\frac{y^2}{8}=1 \quad ...(1)$$ $$\frac{x^2}{6}+\frac{y^2}{9}=1 \quad ...(2)$$ Let $u=x^2$, $v=y^2$. Then $$\frac{u}{12}+\frac{v}{8}=1 \implies 2u+3v=24,$$ $$\frac{u}{6}+\frac{v}{9}=1 \implies 3u+2v=18.$$ Solve: From $$2u+3v=24,$$ $$3u+2v=18.$$ Multiply first by $3$ and second by $2$: $$6u+9v=72,$$ $$6u+4v=36.$$ Subtract: $$5v=36 \implies v=\frac{36}{5}.$$ Then $$3u+2\cdot \frac{36}{5}=18 \implies 3u+\frac{72}{5}=18= rac{90}{5} \implies 3u=\frac{18}{5} \implies u=\frac{6}{5}. $$ Thus, $$x^2=\frac{6}{5}, \qquad y^2=\frac{36}{5}.$$ So the four intersection points are $$\left(\pm \sqrt{\frac65}, \pm \frac{6}{\sqrt5}\right).$$ These form a rectangle with vertices $A,B,C,D$. --- 5. **Find area of quadrilateral $ABCD$** The side lengths of the rectangle are $$2\sqrt{\frac65} \quad \text{and} \quad 2\cdot \frac{6}{\sqrt5} = \frac{12}{\sqrt5}.$$ Therefore area is $$\text{Area}=\left(2\sqrt{\frac65}\right)\left(\frac{12}{\sqrt5}\right).$$ Now, $$\sqrt{\frac65}=\frac{\sqrt6}{\sqrt5},$$ so $$\text{Area}=2\cdot \frac{\sqrt6}{\sqrt5}\cdot \frac{12}{\sqrt5} =\frac{24\sqrt6}{5}. $$ --- 6. **Match with options** $$\boxed{\frac{24\sqrt6}{5}}$$ This is **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer is also **A**, so they agree.

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View written solutionFree

Correct answer: A

Write eccentricity relations for both ellipses

For $E_1: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$ , its eccentricity is $e_1=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{3}}.$ So,

\implies \frac{b^2}{a^2}=\frac{2}{3} \implies b^2=\frac{2a^2}{3}. $$ For $E_2: \dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ with $A<B$, the major axis is along $y$. Hence its eccentricity is $$e_2=\sqrt{1-\frac{A^2}{B^2}}=\frac{1}{\sqrt{3}}.$$ Thus, $$1-\frac{A^2}{B^2}=\frac{1}{3} \implies \frac{A^2}{B^2}=\frac{2}{3} \implies A^2=\frac{2B^2}{3}. $$ --- 2. **Use distance between foci of $E_1$** For $E_1$, focal distance is $2c=4$, so $$c=2.$$ But for an ellipse, $$c=ae= a\cdot \frac{1}{\sqrt{3}}.$$ Therefore, $$\frac{a}{\sqrt{3}}=2 \implies a=2\sqrt{3}. $$ Hence, $$a^2=12, \qquad b^2=\frac{2}{3}a^2=8. $$ So $$E_1:\frac{x^2}{12}+\frac{y^2}{8}=1.$$ --- 3. **Use product of latus recta** Length of latus rectum of $E_1$ is $$L_1=\frac{2b^2}{a}=\frac{2\cdot 8}{2\sqrt3}=\frac{8}{\sqrt3}. $$ Given $$L_1L_2=\frac{32}{\sqrt3},$$ so $$\frac{8}{\sqrt3}L_2=\frac{32}{\sqrt3} \implies L_2=4.$$ For $E_2$ (major axis along $y$), latus rectum length is $$L_2=\frac{2A^2}{B}=4.$$ Using $A^2=\dfrac{2B^2}{3}$, $$\frac{2\left(\frac{2B^2}{3}\right)}{B}=4 \implies \frac{4B}{3}=4 \implies B=3. $$ Then $$A^2=\frac{2B^2}{3}=\frac{2\cdot 9}{3}=6. $$ So $$E_2:\frac{x^2}{6}+\frac{y^2}{9}=1.$$ --- 4. **Find intersection points of the two ellipses** We solve $$\frac{x^2}{12}+\frac{y^2}{8}=1 \quad ...(1)$$ $$\frac{x^2}{6}+\frac{y^2}{9}=1 \quad ...(2)$$ Let $u=x^2$, $v=y^2$. Then $$\frac{u}{12}+\frac{v}{8}=1 \implies 2u+3v=24,$$ $$\frac{u}{6}+\frac{v}{9}=1 \implies 3u+2v=18.$$ Solve: From $$2u+3v=24,$$ $$3u+2v=18.$$ Multiply first by $3$ and second by $2$: $$6u+9v=72,$$ $$6u+4v=36.$$ Subtract: $$5v=36 \implies v=\frac{36}{5}.$$ Then $$3u+2\cdot \frac{36}{5}=18 \implies 3u+\frac{72}{5}=18= rac{90}{5} \implies 3u=\frac{18}{5} \implies u=\frac{6}{5}. $$ Thus, $$x^2=\frac{6}{5}, \qquad y^2=\frac{36}{5}.$$ So the four intersection points are $$\left(\pm \sqrt{\frac65}, \pm \frac{6}{\sqrt5}\right).$$ These form a rectangle with vertices $A,B,C,D$. --- 5. **Find area of quadrilateral $ABCD$** The side lengths of the rectangle are $$2\sqrt{\frac65} \quad \text{and} \quad 2\cdot \frac{6}{\sqrt5} = \frac{12}{\sqrt5}.$$ Therefore area is $$\text{Area}=\left(2\sqrt{\frac65}\right)\left(\frac{12}{\sqrt5}\right).$$ Now, $$\sqrt{\frac65}=\frac{\sqrt6}{\sqrt5},$$ so $$\text{Area}=2\cdot \frac{\sqrt6}{\sqrt5}\cdot \frac{12}{\sqrt5} =\frac{24\sqrt6}{5}. $$ --- 6. **Match with options** $$\boxed{\frac{24\sqrt6}{5}}$$ This is **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer is also **A**, so they agree.

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