If 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

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10-2. Parabola, Ellipse, Hyperbola

hard

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............................

$432$

$456$

$123$

$789$

(JEE MAIN-2024)

Putting $y^2=3 x^2$ in both the conics

We get $x^2=b$ and $\frac{b}{16}+\frac{3}{b}=1$

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

coincide)

$\therefore \mathrm{b}=12$

Hence points of intersection are

$( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432$

Standard 11

Mathematics

hard

hard

(JEE MAIN-2021)

normal

easy

hard

(IIT-1997)