Question

A tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at any point $P$ meets the line $x=0$ at a point $Q$. Let $R$ be the image of $Q$ in the line $y=x$. Then the circle whose extremities of a diameter are $Q$ and $R$ passes through a fixed point. The fixed point is

• A. $(3,0)$
• B. $(5,0)$
• C. $(0,0)$
• D. $(4,0)$

Answer 1

Answer: (C)

The equation of the tangent to the ellipse at $P(5\cos \theta ,4\sin \theta )$ is $\frac{x\cos \theta }{5}+\frac{y\sin \theta }{4}=1$
It meets the line $x=0$ at $Q(0,4\mathrm{cosec}\theta )$
The image of $Q$ on the line $y=x$ is $R(4\mathrm{cosec}\theta ,0)$
Therefore, the equation of the circle is
$x(x-4\mathrm{cosec}\theta )+y(y-\mathrm{cosec}\theta )=0$
i.e., $x^2+y^2-4(x+y)\mathrm{cosec}\theta =0$
Therefore, each number of the family passes through the intersection of $x^2+y^2=0$ and $x+y=0$, i.e., the point $(0,0)$.