Question

Point $(\sqrt{5},\frac{4}{\sqrt{5}})$ lies _____ the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$.

• A. inside
• B. outside
• C. on
• D. None of the above

Answer 1

The correct option is C on

The point $(x,y)$ lies inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if it satisfies the following condition i.e.

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

To check $(\sqrt{5},\frac{4}{\sqrt{5}})$ lies in the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$

Here $a=2$ and $b=5$

Take $(x,y)=(\sqrt{5},\frac{4}{\sqrt{5}})$

$\therefore$ $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{(\sqrt{5}{)}^2}{4}+\frac{(\frac{4}{\sqrt{5}}{)}^2}{25}=\frac{5}{4}+\frac{16}{5\times 25}=\frac{689}{600}<1$

Hence, $(\sqrt{5},\frac{4}{\sqrt{5}})$ lies in the given ellipse.