Question

The distance of the point $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$ from the centre is $2,$ if:

• A. $\theta =\frac{\pi }{4}$
• B. $\theta =\frac{3\pi }{2}$
• C. $\theta =\frac{5\pi }{2}$
• D. $\theta =\frac{-\pi }{4}$

Answer 1

Answer: (A), (D)

The correct options are
A $\theta =\frac{\pi }{4}$
D $\theta =\frac{-\pi }{4}$

Given, equation of ellipse as $\frac{x^2}{6}+\frac{y^2}{2}=1$(where length of major axis$=\sqrt{6}$,$b=$length of minor axis$=$ $\sqrt{2}$.)
center of ellipse is (0,0) which is parallel to horizontal axis and with eccentricity 'e'.
Point on the ellipse $P(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$
Distance from center to point P, $OP=2$(Given)
$\Rightarrow \sqrt{\left(\right(\sqrt{6}cos\theta -0{)}^2+(\sqrt{2}sin\theta -0{)}^2)}=2$
$\Rightarrow \sqrt{(6co{s}^2\theta +2{\sin }^2\theta )}=2$
$\Rightarrow 6co{s}^2\theta +2{\sin }^2\theta =4$
$\Rightarrow 3co{s}^2\theta +{\sin }^2\theta =2$
$\Rightarrow 2co{s}^2\theta +1=2$(since $co{s}^2\theta +si{n}^2\theta =1$)
$\Rightarrow cos\theta =\frac{\pm 1}{\sqrt{2}}$
$\theta =\frac{\pi }{4}or\frac{-\pi }{4}$