Question

If the chords of contact of tangents from two points $(-4,2)$ and $(2,1)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are at right angle, then the eccentricity of the hyperbola, is

• A. $\frac{\sqrt{7}}{2}$
• B. $\sqrt{\frac{5}{3}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

Equation of chord of contact with respect to point $(-4,2)$ is
$\frac{-4x}{a^2}-\frac{2y}{b^2}=1$ and with respect to point $(2,1)$ is $\frac{2x}{a^2}-\frac{y}{b^2}=1$.
Now, according to given condition,
$\left(\frac{\frac{4}{a^2}}{\frac{-2}{b^2}}\right)\times \left(\frac{\frac{-2}{a^2}}{\frac{-1}{b^2}}\right)=-1\Rightarrow \frac{b^4}{a^4}=\frac{1}{4}\Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$