Question

Number of distinct normals, that can be drawn to the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$ from the point $P(0,6)$ is

• A. one
• B. two
• C. three
• D. four

Answer 1

The correct option is C three

Given equation of ellipse is
$\frac{x^2}{169}+\frac{y^2}{25}=1$
Equation of normal at the point $(13\cos \theta ,5\sin \theta )$ is
$\frac{13x}{\cos \theta }-\frac{5y}{\sin \theta }=144$
If it passes through $(0,6),$ then
$\Rightarrow 15+72\sin \theta =0$
$\Rightarrow \sin \theta =-\frac{5}{24}$
$\exists$ two values of theta $\in [0,2\pi ]$ for which $\sin \theta =-\frac{5}{24}$
and $x=0$ will also pass through $(0,6)$
Hence, three normals can be drawn from $(0,6).$