Question:
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:
Correct Option: 1
Solution:
$y^{2}=3 x^{2}$
and $x^{2}+y^{2}=4 b$
Solve both we get
so $\quad x^{2}=b$
$\frac{x^{2}}{16}+\frac{3 x^{2}}{b^{2}}=1$
$\frac{b}{16}+\frac{3}{b}=1$
$b^{2}-16 b+48=0$
$(b-12)(b-4)=0$
$\mathrm{b}=12, \mathrm{~b}>4$
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