Question

The line $5x-3y=8\sqrt{2}$ is a normal to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. If $\theta$ be the eccentric angle of the foot of this normal , then '$\theta$' is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{4}$

From the equation of ellipse given, we have; $a=5,b=3$

Therefore any point on the ellipse $(5cos\theta ,3sin\theta ).$

Normal at this point to the given ellipse is

$5xsec\theta -3ycosec\theta =25-9=16------\left(1\right)$

Also the equation of given normal is $5x-3y=8\sqrt{2}-----\left(2\right)$

Also the equation of the normal $\left(1\right)$ and $\left(2\right)$,

$⟹\frac{5sec\theta }{5}=\frac{3cosec\theta }{3}=\frac{16}{8\sqrt{2}}$

$⟹sin\theta =cos\theta =\frac{1}{\sqrt{2}}$

$⟹\theta =\frac{\Pi }{4}$

Option (c) is correct.