Question

The focal distances of the point $\left(\dfrac{4}{\sqrt{5}}, \dfrac{3}{\sqrt{5}}\right)$ on the ellipse $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ are

• A. $\dfrac{10}{3}, \dfrac{2}{3}$
• B. $3, 1$
• C. $\dfrac{13}{3}, \dfrac{5}{3}$
• D. $4, 2$

Hint:

Use focal length $c$ and distance formula to compute distances from given point to ellipse foci.

Answer 1

The Correct Option is D

Ellipse: $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ ⇒ $a^2 = 9$, $b^2 = 4$ ⇒ $c = \sqrt{a^2 - b^2} = \sqrt{5}$
Foci: $(\pm \sqrt{5}, 0)$
Point $P = \left(\dfrac{4}{\sqrt{5}}, \dfrac{3}{\sqrt{5}}\right)$
Distance to $(\sqrt{5}, 0) = \sqrt{\left(\dfrac{4}{\sqrt{5}} - \sqrt{5}\right)^2 + \left(\dfrac{3}{\sqrt{5}}\right)^2} = \sqrt{4}$
Distance to $(-\sqrt{5}, 0) = \sqrt{\left(\dfrac{4}{\sqrt{5}} + \sqrt{5}\right)^2 + \left(\dfrac{3}{\sqrt{5}}\right)^2} = \sqrt{16} = 4$
Hence distances are $4$ and $2$