If the point of intersections of the ellipse x^216+y^2b^2=1 and the circle x^2+y^2=4b, b4…

Question

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

Accepted Answer

From the circle, $x^2+3x^2=4b\Rightarrow x^2=b$. Substituting $y^2=3x^2$ into the ellipse: $\dfrac{x^2}{16}+\dfrac{3x^2}{b^2}=1\Rightarrow x^2=\dfrac{16b^2}{b^2+48}$.

Equating: $b=\dfrac{16b^2}{b^2+48}\Rightarrow b^2+48=16b\Rightarrow b^2-16b+48=0\Rightarrow b=12	ext{ or }4$.

Since $b>4$, $b=12$.

Solutions · 1