Question

Find the equation of the ellipse with its centre $(1,2)$, focus at $(6,2)$ and passing through the point $(4,6)$.

Answer 1

Given that the centre of ellipse is $(1,2)$, focus is $(6,2)$ and passes through $(4,6)$

$\therefore \frac{{\left(x-1\right)}^2}{a^2}+\frac{{\left(y-2\right)}^2}{b^2}=1$

This is the equation of ellipse

.Now,

Distance between focus and centre is $c$

So, $c=\sqrt{{\left(1-6\right)}^2+{\left(2-2\right)}^2}=5$

So, we have $a^2-b^2=25----\left(1\right)$

As $4(4,6)$ lies on the ellipse, we have

$\frac{{\left(4-1\right)}^2}{a^2}+\frac{{\left(6-2\right)}^2}{b^2}=1$

$\Rightarrow 9b^2+16a^2=a^2{b}^2----\left(2\right)$

On solving $\left(1\right)$ and $\left(2\right)$ we get,

$9b^2+400-16a^2=25b^2+b^2$

$\Rightarrow b^4=400$

$\Rightarrow b^4-400=0$

$\Rightarrow \left(b^2-20\right)\left(b^2+20\right)=0$

$\Rightarrow b^2=20or{b}^2=-20\left(notexist\right)$

Now from $\left(1\right)$ we get

$a^2=20+25=45$

The equation of the ellipse is

$\frac{{\left(x-1\right)}^2}{45}+\frac{{\left(y-2\right)}^2}{20}=1$

Hence, we find the required answer.