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 $Q(0,4 \operatorname{cosec} \theta)$
The image of $Q$ on the line $y=x$ is $R(4 \operatorname{cosec} \theta, 0)$
Therefore, the equation of the circle is
$x(x-4 \operatorname{cosec} \theta)+y(y-\operatorname{cosec} \theta)=0$
i.e., $\quad x^{2}+y^{2}-4(x+y) \operatorname{cosec} \theta=0$
Therefore, each number of the family passes through the intersection of $x^{2}+y^{2}=0$ and $x+y=0$, i.e., the point $(0,0)$.