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Q. 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

Conic Sections

Solution:

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