If point of intersections of ellipse frac{ x ^{2}}{16}+frac{ y ^{2}}{ b ^{2}}=1 and circle x ^

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10-2. Parabola, Ellipse, Hyperbola

hard

If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:

A

$12$

B

$5$

C

$6$

D

$10$

(JEE MAIN-2021)

Solution

$y^{2}=3 x^{2}$

And $x^{2}+y^{2}=4 b$

Solve both we get

So $x^{2}=b$

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

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

$b^{2}-16 b+48=0$

$(b-12)(b-4)=0$

$b=12, b > 4$

Standard 11

Mathematics

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