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 $P$ intersect the curve ar point $Q$ and $R$. If the product $PQ.PR$ is independent of the slope of the line, then the curve is

• A. An ellipse
• B. A hyperbola
• C. A circle
• D. None of these

Answer 1

Answer: (C)

A circle

Let the coordinates of point$P$ be $(x_1,y_1).$

Equation of any line through $P$ can be written as $\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r$ ...(1)

$\Rightarrow x=x_1+r\cos \theta ,y=y_1+r\sin \theta .$

Coordinates of any point an (1) is of the form $(x_1+r\cos \theta ,y_1+r\sin \theta ).$

This point will lie on ${ax}^2+2hxy+{by}^2=1$ if

$a{(x_1+r\cos \theta )}^2+2h(x_1+r\cos \theta )(y_1+r\sin \theta )+b{(y_1+r\sin \theta )}^2-1=0$

$\Rightarrow r^2(a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta )+2[x_1(a\cos \theta +h\sin \theta )+y_1(h\cos \theta +b\sin \theta )]$

$+{ax}_1^2+2{hx}_1{y}_1+{by}_1^2-1=0$ ...(2)

Let $PQ=r_1$ and $PR=r_2.$

Then $r_1,r_2$ are the roots of (2).

$\therefore PQ:PR=r_1{r}_2=\frac{{ax}_1^2+2{hx}_1{y}_1+{by}_1^2-1}{a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta }.$

We know rewrite the denominator.

We have$D=a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta .$

$=\frac{1}{2}[(a+b)+(a-b)\cos 2\theta ]+h\sin 2\theta$

$=\frac{a+b}{2}+\frac{1}{2}(a-b)\cos 2\theta +h\sin 2\theta$

Put $\frac{1}{2}(a-b)=k\sin \alpha ,h=k\cos \alpha .$

$\Rightarrow k=\sqrt{{\left(\frac{a+b}{2}\right)}^2+h^2}$ and $\tan \alpha =\frac{a-b}{2h}$

$\therefore D=\frac{1}{2}(a+b)+\sqrt{{\left(\frac{a-b}{2}\right)}^2+h^2}\sin (2\theta +\alpha )$

Thus, $PQ.PR=\frac{{ax}_1^2+2{hx}_1{y}_1+{by}_1^2-1}{\frac{1}{2}(a+b)+\sqrt{{\left(\frac{a-b}{2}\right)}^2+h^2}\sin (2\theta +\alpha )}$

For this to be independent of $\theta$ we must have ${\left(\frac{a-b}{2}\right)}^2+h^2=0\Rightarrow a=b$ and $n=0.$

But this to be condition for the given curve to represent a circle.