Question

Find the point on the ellipse $\frac{x^2}{4}+\frac{y^2}{36}=1$ whose eccentric angle is $\frac{5\pi }{4}$ radian.

• A. $P(\sqrt{2},-3\sqrt{2})$
• B. $P(\sqrt{2},3\sqrt{2})$
• C. $P(-\sqrt{2},3\sqrt{2})$
• D. $P(-\sqrt{2},-3\sqrt{2})$

Answer 1

The correct option is D $P(-\sqrt{2},-3\sqrt{2})$

Given: $\frac{x^2}{4}+\frac{y^2}{36}=1$ and eccentric angle is $\frac{5\pi }{4}$ radian.

To Find: Point on the ellipse

Step-$1:$ Compare the given equation with standard form of ellipse.

Step-$2:$ Use the obtained values to find the coordinates of the point.

$a=2,b=6$

If $P$ is any point on the ellipse, then coordinates of $P$ are $(a\cos \theta ,b\sin \theta )$.

Eccentric angle is $\theta =\frac{5\pi }{4}$.

$\Rightarrow$ The point is: $P(2\cos \frac{5\pi }{4},6\sin \frac{5\pi }{4})$

$\Rightarrow P(2\times -\frac{1}{\sqrt{2}},6\times -\frac{1}{\sqrt{2}})$

$\Rightarrow P(-\sqrt{2},-3\sqrt{2})$