Question

The distances from the foci to a point $P(x_1,y_1)$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ are :

• A. $4\pm \frac{2}{3}{y}_1$
• B. $5\pm \frac{4}{5}{y}_1$
• C. $5\pm \frac{4}{5}{x}_1$
• D. $4\pm \frac{2}{3}{x}_1$

Answer 1

The correct option is B $5\pm \frac{4}{5}{y}_1$

The equation of the ellipse with center at origin and major axis parallel to the y-axis is given by:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Here $b=3$ and $a=5$

Therefore eccentricity $e=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{25-16}}{5}=\frac{4}{5}$

The focus of the ellipse is given by $S=(\pm ae,0)$

If S is a focus of the ellipse, P any point on the ellipse and if M is the foot of the perpendicular drawn from P to the directrix, then we know $SP=e.PM$

Here $P=(x_1,y_1)$.

The equation of the directrices is given by $y=\pm \frac{b}{e}$

$\Rightarrow$ $y\pm (5\times \frac{5}{4})=0$

$\Rightarrow$ $4y\pm 25=0$ ----------(1)

Therefore PM = Perpendicular distance of $(x_1,y_1)$ from (1) = $\left|\frac{0+4y_1\pm 25}{\sqrt{0+16}}\right|=\frac{25}{4}\pm y_1$

Now $SP=e.PM=\frac{4}{5}.(\frac{25}{4}\pm y_1)=5\pm \frac{4}{5}{y}_1$