Question:
If the point of intersections of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=4 \mathrm{~b}, \mathrm{~b}>4$
lie on the curve $y^{2}=3 x^{2}$, then $b$ is equal to:
Correct Option: 1
Solution:
$y^{2}=3 x^{2}$
and $x^{2}+y^{2}=4 b$
Solve both we get
so $x^{2}=b$
$\frac{x^{2}}{16}+\frac{3 x^{2}}{b^{2}}=1$
$\frac{b}{16}+\frac{3}{b}=1$
$b^{2}-16 b+48=0$
$(b-12)(b-4)=0$
$b=12, b>4$