Question

The distances from the foci of $P(x_{1},y_{1})$ on the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$are

• A. $4 \pm \frac{5}{4}y_{1}$
• B. $5 \pm \frac{4}{5}x_{1}$
• C. $5 \pm \frac{4}{5}y_{1}$
• D. None

Answer 1

Answer: (C)

$5 \pm \frac{4}{5}y_{1}$

Answer 2

For the given ellipse $b > a,$ so the two foci lie on y-axis and their coordinates are $(0, \pm be)$,

Where $b = 5,a = 3$. So $e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$

The focal distances of a point $(x_{1},y_{1})$ on the ellipse$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, Where $b^{2} > a^{2}$are given by $b \pm ey_{1}$. So, Required distances are $b \pm ey_{1} = 5 \pm \frac{4}{5}y_{1}$