Question

In previous question, we have shown that the locus of the feet of the perpendicular draw from the focus of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ upon ay tangent is its auxiliary circle $x^2+y^2=a^2$ then the product of these perpendiculars from the focusupon ay tangent is _____

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let slope of the variable tangent on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be m.

Equation of the tangent $y=mx\pm \sqrt{a^2{m}^2-b^2}$

Let's take the equation of tangent be $mx-y+\sqrt{a^2{m}^2-b^2}=0$

Product of foot of perpendiculars from both the focus is

$p_1.p_2$ = $|\frac{mae+\sqrt{a^2{m}^2-b^2}}{\sqrt{1+m^2}}\times \frac{(-mae+\sqrt{a^2{m}^2-b^2})}{\sqrt{1+m^2}}|$

$p_1.p_2$ = $\left|\frac{(a^2{m}^2-b^2)-m^2\left(a^2{e}^2\right)}{1+m^2}\right|$

=$\left|\frac{a^2{m}^2-b^2-a^2{m}^2-b^2{m}^2}{1+m^2}\right|$

=$\left|\frac{-b^2(1-m^2)}{(1+m^2)}\right|$

=$b^2$

Similarly ,we can show it for other tangent

$y=mx-\sqrt{a^2{m}^2-b^2}$ too

product of the perpendicular on the tangent is constant which is $b^2$