Question

If tangents $PQ$ and $PR$ are drawn from a point on the circle $x^2+y^2=16$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{12}=1(a<3)$, so that the fourth vertex $S$ of the parallelogram $PQSR$ lies on the circumcircle of the triangle $PQR$, then the eccentricity of the ellipse is

• A. $\frac{2}{3}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{\sqrt{2}}{3}$

Answer 1

The correct option is B $\sqrt{\frac{2}{3}}$

concyclic parallelogram is a rectangle .
$\angle QPR=90°\Rightarrow P$ lies on the director circle of the ellipse.
$\Rightarrow a^2+12=16\Rightarrow a^2=4e^2=1-\frac{a^2}{b^2}=\frac{2}{3}e=\sqrt{\frac{2}{3}}$