Question

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

• A. $5$
• B. $6$
• C. $12$
• D. $10$

Answer 1

The correct option is C

$12$

Solve for b

$\left(\frac{x^2}{16}\right)+\left(\frac{y^2}{b^2}\right)=1.....\left(1\right)$

$x^2+y^2=4b...\left(2\right)$

$y^2=3x^2\dots \left(3\right)$

From eqns. $\left(2\right)$ and$\left(3\right),x^2=b$ and $y^2=3b$

From eqn. (1), $\left(\frac{b}{16}\right)+(\frac{3b}{b^2})=1$

$b^2+48=16b$

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

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

but, $b=4$ is not possible,

$b=12$

Hence, option C is the correct option.