Question

If the distance of a point on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ from the centre is $2$, then the eccentric angle is

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

Answer 1

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

Given ellipse equation as $\frac{x^2}{6}+\frac{y^2}{2}=1$

$\Rightarrow$ Length of major axis,a=$\sqrt{6}$ and length of minor axis,b=$\sqrt{2}$.

Any point on the ellipse will be of form $P(acos\theta ,bsin\theta )$.

Center of given ellipse is,C(0,0).

Distance of point P from center is 2(Given).

$\Rightarrow CP=2$

But,$CP=\sqrt{(acos\theta -0{)}^2+(bsin\theta -0{)}^2}$

$\Rightarrow 2=\sqrt{\left(\right(acos\theta -0{)}^2+(bsin\theta -0{)}^2)}$

$\Rightarrow 4=a^2(cos\theta {)}^2+b^2(sin\theta {)}^2$

$\Rightarrow 4=6co{s}^2\theta +2si{n}^2\theta$

$\Rightarrow 4=2+4co{s}^2\theta$

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

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