Question

If tangents $PQ$ and $PR$ are drawn from a point on the circle $x^{2}+y^{2}=25$ to the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1,(b<4)$, so that the fourth vertex $S$ of parallelogram PQSR lies on the circumcircle of triangle $PQR$, then the eccentricity of the ellipse is

• A. $\sqrt{5} / 4$
• B. $\sqrt{7} / 3$
• C. $\sqrt{7} / 4$
• D. $\sqrt{5} / 3$

Answer 1

Answer: (C)

A cyclic parallelogram will be a rectangle or a square.
So, $\angle QPR =90^{\circ}$.
Therefore, P lies on the director circle of the ellipse
$\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$
Hence, $x^{2}+y^{2}=25$
is the director circle of $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b }=1 .$
Then, $16+ b ^{2}=25$
or $\quad b^{2}=9 $
or $a^{2}\left(1-e^{2}\right)=9$
or $1- e ^{2}=\frac{9}{16} $
or $e ^{2}=\frac{7}{16}$
or $e =\frac{\sqrt{7}}{4}$