Question

A variable line $x\cos \alpha +t\sin \alpha =p$ which is a chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(b>a)$ subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius is:

• A. $\frac{ab}{\sqrt{a^2+b^2}}$
• B. $\frac{ab}{\sqrt{b^2-a^2}}$
• C. $\frac{ab}{\sqrt{a^2-b^2}}$
• D. $\frac{ab}{2\sqrt{b^2-a^2}}$

Answer 1

Answer: (B)

$\frac{ab}{\sqrt{b^2-a^2}}$

Given

$x\cos \alpha +y\sin \alpha =P$ is a chord of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Eq${}^n$ of circle centre $(0,0)$ is

$x^2+y^2=p^2$

$(p\cos \alpha ,P\sin \alpha )$

$T:x\times P\cos \alpha +y\times P\sin \alpha =p^2$

$=x\cos \alpha +y\sin \alpha =P$ is tangent to $x^2+y^2=p^2$

By the concept of of How ugerisation

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=(1{)}^2$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}={\left(\frac{x\cos \alpha +y\sin \alpha }{P}\right)}^2$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos x}{P^2}$

$=x^2(\frac{1}{a^2}-\frac{{\cos }^2\alpha }{P^2})y^2(\frac{1}{b^2}-\frac{{\sin }^2\alpha }{p^2})+2xy(\frac{\cos \alpha \sin \alpha }{p^2}=0)$

$OA\perp OB$

$\therefore \frac{1}{a^2}-\frac{{\cos }^2\alpha }{P^2}\frac{-1}{b^2}\frac{-{\sin }^2\alpha }{P^2}=0$

$=\frac{b^2-a^2}{a^2{b}^2}=\frac{1}{p^2}$

$=\frac{ab}{\sqrt{b^2-a^2}}$