Question

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

Answer 1

$x+4y-10=0$

$\Rightarrow y=\frac{-x}{4}+\frac{5}{2}$ .......$\left(1\right)$

Let equation of the ellipse be

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

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

By $\left(1\right)$ we get

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

$\Rightarrow 100=a^2+16b^2$ ........$\left(3\right)$

$(4,-1)$ lies on $\left(3\right)$

$\Rightarrow \frac{16}{a^2}+\frac{1}{b^2}=1$ .....$\left(4\right)$

On solving $\left(3\right)$ and $\left(4\right)$ we get

$\left(3\right)\Rightarrow a^2+16b^2=100$

$\left(4\right)\Rightarrow a^2+16b^2=a^2{b}^2$

From $\left(3\right)$ from $\left(4\right)$ we get

$a^2{b}^2=100$

$\therefore a^2=20$ and $b^2=5$

$\frac{x^2}{20}+\frac{y^2}{5}=1$ is the required equation.