Question

If a variable line $x\cos \alpha +y\sin \alpha =p$ 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 center 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. None of these

Answer 1

Answer: (B)

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

Since $x\cos \alpha +y\sin \alpha =p$ subtends a right angle at the center $(0,0)$ therefore

making equation hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ homogeneous with the help of $x\cos \alpha +y\sin \alpha =p$

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

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

$\Rightarrow$ coefficients of $x^2+$ coefficients of $y^2=0$

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

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

Since $p$ is also the length of the perpendicular from $(0,0)$ to the line $x\cos \alpha +y\sin \alpha =p$

$\therefore$ Radius of the circle $=p=\frac{ab}{\sqrt{b^2-a^2}}$