Question

Consider a curve $a{x}^2+2hxy+b{y}^2=1$ and a point $P$ not on the curve. A line drawn from the point $P$ intersects the curve at points $Q$ and $R$. If the product $PQ$. $PR$ is independent of the slope of the line, then show that the curve is a circle.

Answer 1

Any line through $P(\alpha ,\beta )$ is
$\frac{x-\alpha }{\cos \theta }=\frac{y-\beta }{\sin \theta }=r$ (say)
It meets the given curve where
${(r\cos \theta +\alpha )}^2+2h(r\cos \theta +\alpha )(r\sin \theta +\beta )+b{(r\sin \theta +\beta )}^2=1$
$r^2(a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta )+r\left(\right)+(a{\alpha }^2+2h\alpha \beta +b{\beta }^2-1)=0$
Above is a quadratic in $r$ and gives two values of $r$ say
$r_1$ and $r_2$ which represent $PQ$ and $PR$
$\therefore PQ.PR=r_1.r_2$
$=\frac{a{\alpha }^2+2h\alpha \beta +b{\beta }^2-1}{a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta }$
Above result will be independent of slop i.e. $\theta$ if $a=b$
and $h=0\because {\cos }^2\theta +{\sin }^2\theta =1$
Hence the given equation of curve becomes a circle as
$a=b$ and $h=0$.