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 ecentricity of the hyperbola is

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

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}$

The chord of contact from $\left(x_1{y}_1\right)$ to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $\frac{x{x}_1}{a^2}-\frac{y{y}_2}{b^2}=1$

Chord of contact from $(-4,2)\Rightarrow \frac{-4x}{a^2}-\frac{2y}{b^2}=1$

$slope\left(m_1\right)=\frac{-2b^2}{a^2}$

Chord of contact from $(2,1)\Rightarrow \frac{2x}{a^2}-\frac{y}{b^2}=1$

$slope\left(m_2\right)=\frac{2b^2}{a^2}$

As both are $\perp ,m_1{m}_2=-1$

$\Rightarrow \left(\frac{{-2b}^2}{a^2}\right)\left(\frac{{2b}^2}{a^2}\right)=-1$

$\Rightarrow \frac{b^4}{a^4}=\frac{1}{4}$

$\Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$

Eccentricity, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$\therefore OptionCiscorrect$