Question

Let two perpendicular chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$ each passing through exactly one of the foci meet at point $P$. If from $P$ two tangents are drawn to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then $\angle QPR=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $2{\tan }^{-1}\frac{b}{a}$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

The equation of two perpendicular chords drawn through each of the foci be $y=m(x-ae)$ and $y=-\frac{1}{m}(x-ae).$

Locus of their point of intersection $P$ is obtained by eliminating the variable $m$. Multiplying the equations of chords, we have $y^2=-(x^2-a^2{e}^2)$ or $x^2+y^2=a^2-b^2$
Above represents the director circle of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which we know is the locus of the point of intersection of perpendicular tangents $QP$ and $QR$.

Hence, $\angle QPR=\frac{\pi }{2}$