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Q.
The eccentric angle of the point where the line, $5 x-3 y=8 \sqrt{2}$ is a normal to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ is
Conic Sections
Solution:
Normal to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ at the point ' $\theta^{\prime}$ is $\frac{5 x}{\cos \theta}-\frac{3 y}{\sin \theta}=16$
Comparing with, $5 x-3 y=8 \sqrt{2} \cos \theta=\sin \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}$