Question

The eccentric angle of the point where the line, $5x-3y=8\sqrt{2}$ is a normal to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ is

• A. $\frac{3\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $ta{n}^{-1}2$

Answer 1

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

The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $P(a\cos \theta ,b\sin \theta )$ is $ax\sec \theta -bycosec\theta =a^2-b^2$

Given,ellipse equation as $\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow$ Length of major axis, $a=5$ and length of minor axis, $b=3$.

$\therefore$The required equation of normal is $5x\sec \theta -3ycosec\theta =5^2-3^2$

$\Rightarrow 5x\sec \theta -3ycosec\theta =16\dots \left(1\right)$

Given normal equation $5x-3y=8\sqrt{2}$

Multiplying both sides with $\sqrt{2}$

$\Rightarrow 5\sqrt{2}x-3\sqrt{2}y=16\dots \left(2\right)$

Comparing equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \sec \theta =\sqrt{2}$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$