The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2-2 y^2-2=0$ to its…
The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2-2 y^2-2=0$ to its asymptotes is :
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{3}{2}$
2
Solution
Equation of hyperbola is
$x^2-2 y^2=2$
$\Rightarrow \quad \frac{x^2}{2}-\frac{y^2}{1}=1$
Here, $a^2=2, b^2=1$
Equation of asymptotes to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } \frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
$\therefore \quad \frac{x}{a}-\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}=0$
Let $P(a \sec \theta, b \tan \theta)$ be any point, then the product of length of perpendiculars
$=\frac{\left[\frac{a \sec \theta}{a}+\frac{b \tan \theta}{b}\right]}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \frac{\left[\frac{a \sec \theta}{a}+\frac{b \tan \theta}{b}\right]}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$=\frac{\sec ^2 \theta-\tan ^2 \theta}{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{1}{\frac{1}{2}+\frac{1}{1}}=\frac{2}{3}$