Question

The product of the perpendiculars from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is equal to $k\frac{a^2{b}^2}{a^2+b^2}$ then $k=$

Answer 1

Given hyperbola$\Rightarrow S\equiv \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\therefore$Asymptotes are $y=\pm \frac{b}{a}x$

$\Rightarrow bx-ay=0$ Or $bx+ay=0$

Let $(x_0,y_0)$ be any point on $S=0$

Perpendicular to $bx-ay=0$

$d_1=\frac{|b{x}_0-a{y}_0|}{\sqrt{a^2+b^2}}$

Perpendicular to $bx+ay=0$

$d_2=\frac{|b{x}_0+a{y}_0|}{\sqrt{a^2+b^2}}$

$\therefore d_1.d_2=\frac{(b^2{x}_0^2-a^2{y}_0^2)}{a^2+b^2}$

But $b^2{x}_0^2-a^2{y}_0^2=a^2{b}^2$

$\therefore d_1.d_2=\frac{a^2{b}^2}{a^2+b^2}$

$\Rightarrow k=1$