Question

The product of the lengths of the perpendiculars from any point on the hyperbola $x^2-2y^2=2$ to its asymptotes is

• A. $2$
• B. $\frac{3}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Given hyperbola is $S\equiv \frac{x^2}{2}-\frac{y^2}{1}=1$
Asymptotes are
$\Rightarrow x-\sqrt{2}y=0$ or $x+\sqrt{2}y=0$

Let $P(x_0,y_0)$ be any point on $S=0$ Length of the perpendicular from $P$ to $x-\sqrt{2}y=0$ is $d_1=\frac{|x_0-{\sqrt{2}y}_0|}{\sqrt{3}}$

Length of the perpendicular from $P$ to $x+\sqrt{2}y=0$ is $d_2=\frac{|x_0+{\sqrt{2}y}_0|}{\sqrt{3}}$
$\therefore d_1\cdot d_2=\frac{(x_0^2-2y_0^2)}{3}$
But $x_0^2-2y_0^2=2$
$\therefore d_1\cdot d_2=\frac{2}{3}$

ALTERNATE ::
Product of length of perpendiculars from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is $\frac{a^2{b}^2}{a^2+b^2}=\frac{2}{3}$