Question

Find the equation of ellipse whose axis are co-ordinate axis and passes through point $(6,2)$ and $(4,3)$.

Answer 1

Standard equation of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It passes through point $(6,2)$
then $\frac{6^2}{a^2}+\frac{2^2}{b^2}=1$
$\Rightarrow \frac{36}{a^2}+\frac{4}{b^2}=\mathrm{1...}\left(i\right)$
It also passes through point $(4,3)$
$\frac{4^2}{a^2}+\frac{3^2}{b^2}=1$
$\Rightarrow \frac{16}{a^2}+\frac{19}{b^2}=\mathrm{1...}\left(ii\right)$
Multiply eqn $\left(i\right)$ by $9$ and eqn $\left(ii\right)$ by $4$ then subtracting
$\frac{324}{a^2}+\frac{36}{b^2}=9$
$\frac{64}{a^2}+\frac{36}{b^2}=4$
$\frac{260}{a^2}=5$
$a^2=\frac{260}{5}=52$
Put the value of $a^2$ in equation $\left(i\right)$,
$\frac{16}{52}+\frac{9}{b^2}=1$
$\Rightarrow \frac{4}{13}+\frac{9}{b^2}=1$
$\Rightarrow \frac{9}{b^2}=1-\frac{4}{13}=\frac{9}{13}$
Thus equation of ellipse is $\frac{x^2}{52}+\frac{y^2}{13}=1$