Question

A point on the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at a distance equal to the mean of the lengths of the semi major axis and semi minor axis from the centre is

• A. $(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{14})$
• B. $(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{7})$
• C. $(\pm \frac{2\sqrt{105}}{7},\pm \frac{3\sqrt{91}}{14})$
• D. $(\pm \frac{2\sqrt{105}}{14},\pm \frac{3\sqrt{91}}{14})$

Answer 1

The correct option is A $(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{14})$

Length of semi major axis and semi minor axis of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ are $4$ and $3$ respectively. So mean of these lengths is $\frac{7}{2}.$

Let the coordinates of any point on the ellipse be $P(4\cos \theta ,3\sin \theta ).$ If the distance of $P$ from the centre$(0,0)$ of the ellipse is $\frac{7}{2},$ then $16{\cos }^2\theta +9{\sin }^2\theta =\frac{49}{4}$
$\Rightarrow 28{\cos }^2\theta =13\Rightarrow \cos \theta =\pm \sqrt{\frac{13}{28}}=\pm \frac{\sqrt{91}}{14}$
and $\sin \theta =\pm \frac{\sqrt{105}}{14}.$

So, the coordinates of the required point are $(\pm \frac{4\sqrt{91}}{14},\pm \frac{3\sqrt{105}}{14})$
or $(\pm \frac{2\sqrt{91}}{7},\pm \frac{3\sqrt{105}}{14})$