Question

Prove that the product of perpendiculars from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is constant and the value is $\frac{a^2{b}^2}{a^2+b^2}$

Answer 1

The equation of asymptotes are $bx+ay=0$ and $bx-ay=0$

Let the point on hyperbola be $(p,q)$

Perpendicular distance from the point $(p,q)$ to the line $bx+ay=0$ is $\frac{bp+ay}{\sqrt{a^2+b^2}}$

Perpendicular distance from point $(p,q)$ to the line $bx-ay=0$ is $\frac{bp-aq}{\sqrt{a^2+b^2}}$

The product of distances will be $\frac{b^2{p}^2-a^2{q}^2}{a^2+b^2}$

The point $(p,q)$ lie on hyperbola , so by substituting we get $b^2{p}^2-a^2{q}^2=a^2{b}^2$

Therefore the product of distances is $\frac{a^2{b}^2}{a^2+b^2}$

Hence proved