• # question_answer 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) $\sqrt{2}$

 Chord of contact of tangent from$(-\,4,2)$the hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$is $\frac{-\,4x}{{{a}^{2}}}-\frac{2y}{{{b}^{2}}}=1$ ? (i) And from (2, 1) is $\frac{2x}{{{a}^{2}}}-\frac{y}{{{b}^{2}}}=1$ ? (ii) Since, chord of contact are perpendicular. $\therefore$$\left( \frac{-\,4}{{{a}^{2}}} \right)\left( \frac{2}{{{a}^{2}}} \right)+\left( \frac{2}{{{b}^{2}}} \right)\left( \frac{1}{{{b}^{2}}} \right)=0$ $\frac{{{a}^{4}}}{{{b}^{4}}}=4$
 $\Rightarrow$   ${{a}^{2}}=2{{b}^{2}}$ $\Rightarrow$   $\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{2}$ $\Rightarrow$   $e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}$ $=\sqrt{1+\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}$