Question

If the extremities of the latus rectum of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is $(\alpha ,\beta )$, then the distance between the point $P(1,1)$ and $(\alpha ,\beta )$, when $\alpha >0$ is/are

• A. $\frac{\sqrt{221}}{25}\text{units}$
• B. $\frac{\sqrt{221}}{5}\text{units}$
• C. $\frac{\sqrt{541}}{5}\text{units}$
• D. $\frac{\sqrt{541}}{25}\text{units}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{\sqrt{221}}{5}\text{units}$
C $\frac{\sqrt{541}}{5}\text{units}$

Given equation of ellipse is
$\frac{x^2}{25}+\frac{y^2}{16}=1$
It is in the form of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a=5$ and $b=4$
Clearly $a>b$
Now,
$e^2=1-\frac{b^2}{a^2}\Rightarrow e^2=1-\frac{16}{25}\Rightarrow e=\frac{3}{5}$

The coordinates of the extremities of the latus rectum are
$(\pm ae,\pm \frac{b^2}{a})$
$\Rightarrow (\alpha ,\beta )=(\pm 3,\pm \frac{16}{5})$
Since $\alpha >0$, then coordinates are $(3,\frac{16}{5})$ and $(3,-\frac{16}{5})$

$\therefore$ Distance between $(1,1)$ and $(3,\frac{16}{5})$
$=\sqrt{(1-3{)}^2+{(1-\frac{16}{5})}^2}=\frac{\sqrt{221}}{5}\text{units}$

And the distance between $(1,1)$ and $(3,-\frac{16}{5})$
$=\sqrt{(1-3{)}^2+{(1+\frac{16}{5})}^2}=\frac{\sqrt{541}}{5}\text{units}$