Ellipse Question 3
Question 3 - 2024 (29 Jan Shift 1)
If the points of intersection of two distinct conics $x^{2}+y^{2}=4 b$ and $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ lie on the curve $y^{2}=3 x^{2}$, then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is
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Answer (432)
Solution
Putting $y^{2}=3 x^{2}$ in both the conics
We get $x^{2}=b$ and $\frac{b}{16}+\frac{3}{b}=1$
$\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{~b}=4$ is rejected because curves coincide $)$
$\therefore \mathrm{b}=12$
Hence points of intersection are
$( \pm \sqrt{12}, \pm 6) \Rightarrow$ area of rectangle $=432$
