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 $b^2=9$
or $a^2\left(1-e^2\right)=9$
or $1-e^2=\frac{9}{16}$
or $e^2=\frac{7}{16}$
or $e=\frac{\sqrt{7}}{4}$