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

Answer 1

Answer: (A)

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

Answer 2

Given ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ i.e. $\frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$.

∴ Lengths of semi-major and semi-minor axes are 4 and 3 respectively. So, the mean of these lengths is 7/2.

Let P (4cos θ, 3 sinθ) be a point on the ellipse at a distance 7/2 from the centre (0, 0).

∴ 16cos²θ + 9sin²θ = $\frac{49}{4}$

⇒ 16cos²θ+ 9 (1 - cos²θ) = $\frac{49}{4}$ ⇒ 28cos²θ = 13

⇒ cosθ = ± $\sqrt{\frac{13}{28}} = \pm \sqrt{\frac{91}{14}}$ and sinθ = ±$\sqrt{\frac{105}{14}}$.

So, the coordinates of the required point are

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