JEE Main 2024 — Ellipse Question with Solution

Given conics are $x^2+y^2=4b \text{and} \frac{x^2}{16}+\frac{y^2}{b^2}=1$.

The intersection point of these conics lie on $y^2=3x^2$.

So, putting ${\mathrm{y}}^2=3{\mathrm{x}}^2$ in both the conics.

$\Rightarrow x^2+3x^2=4b, \frac{x^2}{16}+\frac{3x^2}{b^2}=1$

$\Rightarrow x^2=b, \frac{b}{16}+\frac{3b}{b^2}=1$

$\Rightarrow \frac{b}{16}+\frac{3}{b}=1$

$\Rightarrow b^2+48=16b$

$\Rightarrow b^2-16b+48=0$

$\Rightarrow b^2-12b-4b+48=0$

$\Rightarrow \left(b-4\right)\left(b-12\right)$

$\Rightarrow b= 4, 12$ ( $b=4$is rejected because curves coincide)

$\Rightarrow \mathrm{b}=12$

$\Rightarrow x=\pm \sqrt{12}$

$\Rightarrow 12+y^2=48$

$\Rightarrow y=\pm 6$

Hence points of intersection are $(\pm \sqrt{12},\pm 6)$.

So, length and breadth of rectangle are $2\sqrt{12} \text{and} 12$. 

So, the area of rectangle is given by, $A=2\sqrt{12}\times 12=48\sqrt{3}$

$\Rightarrow 3\sqrt{3}A=48\sqrt{3}\times 3\sqrt{3}$

$\Rightarrow 3\sqrt{3}A=432$