Question

If a variable straight line $x\cos \alpha +y\sin \alpha =p$ which is a chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(b>0)$ subtend a right angle at the centre of the hyperbola, then it always touches a fixed circle whose:

• A. radius is $\frac{ab}{\sqrt{b-2a}}$
• B. radius is $\frac{ab}{\sqrt{b^2-a^2}}$
• C. centre is $(0,0)$
• D. centre is at the centre of the hyperbola

Answer 1

Answer: (B), (C), (D)

The correct options are
B radius is $\frac{ab}{\sqrt{b^2-a^2}}$
C centre is $(0,0)$
D centre is at the centre of the hyperbola

Equation of the pair of straight lines passing through the origin
(centre of the hyperbola) and points of intersection of the variable
chord and the hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}-{\left\{\frac{x\cos \alpha +y\sin \alpha }{p}\right\}}^2=0$
They are at right angles if

$[\frac{1}{a^2}-\frac{{\cos }^2\alpha }{p^2}]-[\frac{1}{b^2}+\frac{{\sin }^2\alpha }{p^2}]=0$
$\Rightarrow$ $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{p^2}\Rightarrow p=\frac{ab}{\sqrt{b^2-a^2}}$
As
p is the length of the perpendicular from the origin on the line
$x\cos \alpha +y\sin \alpha =p$, line touches the circle with centre
at the origin and radius equal to $\frac{ab}{\sqrt{b^2-a^2}}$