Question

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

• A. $\frac{ab}{a^2+b^2}$
• B. $\frac{ab}{\sqrt{b^2-a^2}}$
• C. $\frac{ab}{\sqrt{b^2+a^2}}$
• D. none of these.

Answer 1

Answer: (B)

If $xcos\alpha +ysin\alpha =p$meets the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$at the points $A,B$, then join equation of $OA,OB$ (where $0$ is the centre of the hyperbola) is given by
$\frac{x^2}{a^2}-\frac{y^2}{b^2}={\left(\frac{xcos\alpha +ysin\alpha }{p}\right)}^2$
i.e., $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{x^{2}co{s}^2\alpha }{p^2}-\frac{y^{2}si{n}^2\alpha }{p^2}-\frac{2xypcos\alpha sin\alpha }{p^2}=0$
Since the chord subtends a right angle at the centre
$\therefore \frac{1}{a^2}-\frac{co{s}^2\alpha }{p^2}-\frac{1}{b^2}-\frac{si{n}^2\alpha }{p^2}=0$
$\Rightarrow \frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{p^2}$
$\Rightarrow p=\frac{ab}{\sqrt{b^2-a^2}}$
Now, $xcos\alpha +ysin\alpha =p$ touches $x^2+y^2=r^2$
if length of the $\perp$ from $\left(0,0\right)$ upon the line = Radius
i.e., if $\frac{p}{\sqrt{co{s}^2\alpha +si{n}^2\alpha }}=r$ i.e., $p=r$
Hence line touches a fixed circle with radius
$\frac{ab}{\sqrt{b^2-a^2}}$