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

Answer 1

Answer: (A)

$\left( \pm \frac{2\sqrt{91}}{7}, \pm \frac{3\sqrt{105}}{14} \right)$

Answer 2

Lengths 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 that the mean of these length is $\frac{7}{2}$. Let the co-ordinates of any point on the ellipse be P (4 cos q, 3 sin q). If the distance of P from the centre O(0, 0) of the ellipse is $\frac{7}{2}$, then 16 cos² q + 9 sin² q = $\frac{49}{4}$.

Ž 28 cos² q = 13 Ž cos q = ± $\sqrt{\frac{13}{28}}$ = ± $\frac{\sqrt{91}}{14}$ and

sin q = ±$\frac{\sqrt{105}}{14}$.

So, the co-ordinates of the required point are

$\left( \pm \frac{4\sqrt{91}}{14}, \pm \frac{3\sqrt{105}}{14} \right)$ i.e. $\left( \pm \frac{2\sqrt{91}}{7}, \pm \frac{3\sqrt{105}}{14} \right)$