Question

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

• A. equation of the curve is a circle with radius $\frac{1}{\sqrt{a}}$
• B. locus of $P$ is a straight line with $x$-intercept $a$
• C. equation of the curve is a circle with radius $\frac{1}{\sqrt{b}}$
• D. locus of $P$ is a straight line with $x$-intercept $b$

Answer 1

Answer: (A), (C)

The correct options are
A equation of the curve is a circle with radius $\frac{1}{\sqrt{a}}$
C equation of the curve is a circle with radius $\frac{1}{\sqrt{b}}$

Let P be $(x_1,y_1)$ and the line through $P(x_1,y_1)$ makes an angle $\theta$ with $x$-axis. Then
$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r$

Coordinates of any point on the line is $(x_1+r\cos \theta ,y_1+r\sin \theta )$
Since the line through $P$ intersects the curve at $Q$ and $R$
$\therefore a(x_1+r\cos \theta {)}^2+2h(x_1+r\cos \theta \left)\right(y_1+r\sin \theta )+b(y_1+r\sin \theta {)}^2=1$

It is a quadratic equation in $r.$
If the roots of the equation are $r_1\text{and}{r}_2,$ then
$r_1\cdot r_2=\frac{a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2-1}{a{\cos }^2\theta +h\sin 2\theta +b{\sin }^2\theta }=PQ\cdot QR$

Since $PQ\times QR$ is independent of the slope, we have
$a=b,h=0$
Then $PQ\times QR=x_1^2+y_1^2-\frac{1}{a}$
and the curve becomes $x^2+y^2=\frac{1}{a}$