Question

Equation of tangent to ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ at point P whose eccentric angle is $\theta ,$ is $\frac{x}{2\sqrt{3}}+\frac{y}{10}=1.$ Then eccentric angle of point P equals:

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

Answer 1

The correct option is D $\frac{\pi }{6}$

Given equation of ellipse is :
$\frac{x^2}{9}+\frac{y^2}{25}=1$
$\text{Here,}a=3\text{and}b=5$
Equation of tangent in terms of $\theta :$
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
$\Rightarrow \frac{x\cos \theta }{3}+\frac{y\sin \theta }{5}=1\cdots \cdots \left(1\right)$
Equation of tangent given in the question: $\frac{x}{2\sqrt{3}}+\frac{y}{10}=1\cdots \cdots \left(2\right)$

Since, equation $\left(1\right)\text{and}\left(2\right)$ are similar, and RHS of both equations are equal, so the coefficient of $x$ and $y$ of both equations are also equal.
$\therefore \frac{\cos \theta }{3}=\frac{1}{2\sqrt{3}}\text{and}\frac{\sin \theta }{5}=\frac{1}{10}$
$\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\text{and}\sin \theta =\frac{1}{2}$
By values of $\sin \theta$ and $\cos \theta$, we can conclude that value of $\theta =\frac{\pi }{6}$
Therefore, the correct answer is option (d).