Question

Find the equation of ellipse in the stranded from, if it passes through the points $(-2,2)$ and $(3,-1)$

Answer 1

We shall take $(0,0)$ as the center.

The equation of the ellipse is -

$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$

At Center, it becomes,

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

The given ellipse passes through $(-2,2)$

$⟹\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$⟹\frac{4}{a^2}+\frac{4}{b^2}=1$ ------ (i)

The given ellipse passes through $(3,-1)$

$⟹\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$⟹\frac{9}{a^2}+\frac{1}{b^2}=1$ ------ (ii)

Let's solve (i) and (ii), so - let $\frac{1}{a^2}=m$ and $\frac{1}{b^2}=n$

The equations hence become -: $4m+4n=1$ and $9m+n=1$

Solving these two, we get

$m=\frac{3}{32}$ and $n=\frac{5}{32}$

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ becomes $\frac{3x^2}{32}+\frac{5y^2}{32}=1$

The equation of the ellipse is, $\frac{3x^2}{32}+\frac{5y^2}{32}=1$

or $3x^2+5y^2=32$