If a variable line $x\cos\alpha+y\sin\alpha=p$ which is a chord of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $(b>a)$ subtend a right angle at the centre of hyperbola,then prove that it always touches a fixed circle whose radius is $\dfrac{ab}{\sqrt{b^2-a^2}}$
$x\cos\alpha+y\sin\alpha=p$ touches(tangent to) a fixed circle $x^2+y^2=p^2$ of radius $p$. So I need to find $p$.
Let $A$,$B$ be the points where the given lines intersects the hyperbola, ie. $\angle AOB=90^\circ$
How do I proceed further and find the radius of the required circle ?
Note: I really do not want to use some existing shortcut formula here
Best Answer
The line equation seems a bit of a distraction. Here's how I'd solve the problem.
Suppose $H$ and $K$ determine a chord of the hyperbola that subtends a right angle at its center (the origin), $O$. We can write $$H = (h \cos\phi, h \sin\phi) \qquad K = (k \sin\phi,-k \cos\phi)$$ where $h:=|OH|$, $k:=|OK|$, and $\phi$ is some arbitrary angle. Define $p := |OP|$, where $P$ is the foot of the altitude from $O$ to $\overleftrightarrow{HK}$. Calculating the area of $\triangle HOK$ in two ways, we have
$$\frac12|OH||OK| = \frac12|OP||HK|\quad\to\quad hk = p\sqrt{h^2+k^2}\quad\to\quad p = \frac{1}{\sqrt{\dfrac{1}{h^2}+\dfrac{1}{k^2}}} \tag{1}$$
Since $H$ and $K$ both lie on the hyperbola, we have $$\begin{align} \frac{h^2}{a^2}\cos^2\phi - \frac{h^2}{b^2}\sin^2\phi &= 1 \tag{2} \\[4pt] \frac{k^2}{a^2}\sin^2\phi - \frac{k^2}{b^2}\cos^2\phi &= 1 \tag{3} \end{align}$$ Therefore, $k^2 (2) + h^2 (3)$ becomes $$\frac{h^2k^2}{a^2}(\cos^2\phi+\sin^2\phi)-\frac{h^2k^2}{b^2}(\sin^2\phi+\cos^2\phi) = h^2 + k^2 \quad\to\quad \frac{1}{a^2}-\frac{1}{b^2} = \frac{1}{h^2} + \frac{1}{k^2} \tag{4} $$ Thus, recalling $(1)$, we have $$p = \frac{1}{\sqrt{\dfrac{1}{a^2}-\dfrac{1}{b^2}}} = \frac{ab}{\sqrt{b^2-a^2}} \tag{5}$$ Since $p$ is the distance from $\overleftrightarrow{HK}$ to the origin, we conclude that the line is always tangent to the circle with the radius given in $(5)$, as desired. $\square$