Question

Find the equation for the ellipse that satisfies the given conditions: Center at $(0,0)$, major axis on the
$y$-axis and passes through the points $(3,2)$ and $(1,6).$

Answer 1

Given: Ellipse has center at $(0,0)$ major axis on the $y$-axis and passes through the points $(3,2)$ and $(1,6).$
$\because$ Major axis is along $y$-axis and center is at the origin.
$\therefore$ Equation of ellipse is of the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1...\left(1\right)$
Points $(3,2)$ and $(1,6)$ will satisfy equation of ellipse.
From point $(3,2)$ and equation $\left(1\right)$
$\Rightarrow \frac{(3{)}^2}{b^2}+\frac{(2{)}^2}{a^2}=1$
$\Rightarrow \frac{9}{b^2}+\frac{4}{a^2}=1...\left(2\right)$
From point $(1,6)$ and equation $\left(1\right)$
$\Rightarrow \frac{(1{)}^2}{b^2}+\frac{(6{)}^2}{a^2}=1$
$\Rightarrow \frac{1}{b^2}+\frac{36}{a^2}=1$
$\Rightarrow \frac{1}{b^2}=1-\frac{36}{a^2}...\left(3\right)$

Now, from equation $\left(2\right)$ and $\left(3\right)$
$\Rightarrow 9(1-\frac{36}{a^2})+\frac{4}{a^2}=1$
$\Rightarrow 9-\frac{324}{a^2}+\frac{4}{a^2}=1$
$\Rightarrow \frac{-324+4}{a^2}=1-9$
$\Rightarrow \frac{-320}{a^2}=-8$
$\Rightarrow a^2=\frac{320}{8}$
$\Rightarrow a^2=40$
Putting value of $a^2=40$ in $\left(3\right),$
i.e., $\frac{1}{b^2}=1-\frac{36}{a^2}$
$\Rightarrow \frac{1}{b^2}=1-\frac{36}{40}$
$\Rightarrow \frac{1}{b^2}=1-\frac{9}{10}$
$\Rightarrow \frac{1}{b^2}=\frac{10-9}{10}$
$\Rightarrow \frac{1}{b^2}=\frac{1}{10}$
$\Rightarrow b^2=10$
Thus, $a^2=40\&b^2=10$
$a^2=40\&b^2=10$
Putting value of $a^2$ and $b^2$ in $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Hence, required equation of ellipse is $\frac{x^2}{10}+\frac{y^2}{40}=1$