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 $x \,cos \,\alpha + y \, sin \,\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{ x \, cos \,\alpha +y \,sin \,\alpha}{p}\right)^{2} $
i.e., $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{ x^{2}cos^{2}\alpha}{p^{2}} -\frac{ y^{2}sin^{2}\alpha}{p^{2}} - \frac{2 \,xy \, p \,cos \,\alpha \, sin \, \alpha}{p^{2}} = 0 $
Since the chord subtends a right angle at the centre
$ \therefore \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}}}$
Now, $x \,cos \,\alpha + y \,sin \,\alpha = p $ touches $x^{2} + y^{2} = r^{2}$
if length of the $\bot$ from $\left(0, 0\right)$ upon the line = Radius
i.e., if $\frac{p}{\sqrt{cos^{2}\alpha+sin^{2}\alpha}} = r$ i.e., $ p=r$
Hence line touches a fixed circle with radius
$ \frac{ab }{\sqrt{b^{2}-a^{2}}}$