Question

An ellipse passes through the point $(4,-1)$ and touches the line $x+4y-10=0$. Find its equation if its axes coincide with co-ordinate axes.

Answer 1

We have,

$\Rightarrow x+4y-10=0\Rightarrow 4y=-x+10\Rightarrow y=\frac{-x}{4}+\frac{5}{2}\dots \dots \left(i\right)$

Let equation or ellipse be,

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

$\Rightarrow y=mx+c$ is tangent to ellipse if $c^2=a^2{m}^2+b^2$,

By equation (i) we get,

$\Rightarrow \frac{25}{4}=\frac{a^2}{16}+b^2$

$\Rightarrow 100=a^2+16b^2\dots \dots \left(iii\right)$

$(4,-1)$ lies on (iii)

$\Rightarrow \frac{16}{a^2}+\frac{1}{b^2}=1\dots \dots \left(iv\right)$

On having $\left(iii\right)$ and $\left(iv\right)$, we get

$a^2=20$ $b^2=5$

i.e. $\frac{x^2}{20}+\frac{y^2}{5}=1$