Question

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

Answer 1

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Since major axis is along $y-$axis and centre is at $(0,0)$

So, required equation of ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ ..........$\left(1\right)$

Given that ellipse passes through point $(3,2)$ and $(1,6)$

Points $(3,2)$ and $(1,6)$ will satisfy equation of ellipse.

Put $x=3$ and $y=2$ in eqn$\left(1\right)$ we get

$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1$

or $\frac{9}{b^2}+\frac{4}{a^2}=1$ ........$\left(2\right)$

Put $x=1$ and $y=6$ in eqn$\left(1\right)$ we get

$\frac{1^2}{b^2}+\frac{6^2}{a^2}=1$

or $\frac{1}{b^2}+\frac{36}{a^2}=1$ ........$\left(3\right)$

Now equations are

$\frac{9}{b^2}+\frac{4}{a^2}=1$ ........$\left(2\right)$

$\frac{1}{b^2}+\frac{36}{a^2}=1$ ........$\left(3\right)$

From $\left(3\right)$

$\frac{1}{b^2}+\frac{36}{a^2}=1$

$\Rightarrow \frac{1}{b^2}=1-\frac{36}{a^2}$

Putting the value of $b^2$ in eqn$\left(2\right)$ we get

$9(1-\frac{36}{a^2})+\frac{4}{a^2}=1$

$\Rightarrow 9\frac{324}{a^2}+\frac{4}{a^2}=1$

$\Rightarrow 9\frac{-320}{a^2}=-8$

$\Rightarrow a^2=\frac{-320}{-8}=40$

Putting the value of $a^2=40$ in eqn$\left(3\right)$ we get

$\frac{1}{b^2}+\frac{36}{a^2}=1$

$\Rightarrow \frac{1}{b^2}=1-\frac{36}{a^2}$

$\Rightarrow \frac{1}{b^2}=1-36\frac{1}{a^2}$

$\Rightarrow \frac{1}{b^2}=1-\frac{36}{40}$

$\Rightarrow \frac{1}{b^2}=\frac{40-36}{40}$

$\Rightarrow \frac{1}{b^2}=\frac{4}{40}=\frac{1}{10}$

$\therefore b^2=10$

Now required equation of ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Putting value of $b^2$ and $a^2$ we get

$\frac{x^2}{10}+\frac{y^2}{40}=1$

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