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 the coordinate axes.

Answer 1

consider equation of ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Putting $(4,-1)$ as it lies on it

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

equation of tangent of ellipse is

$y=mx\pm \sqrt{a^2{m}^2+b^2}$

compare it with given one

$y=\frac{x}{4}+\frac{5}{2}$

$\frac{5}{2}=\sqrt{a^2{m}^2+b^2}$where $m=\frac{1}{4}$

$\frac{25}{4}=\frac{a^2}{16}+b^2.....\left(2\right)$

$\frac{16}{2}=1-\frac{1}{b^2}$ from (1)

$\frac{a^2}{16}=\frac{1}{1-\frac{1}{b^2}}=\frac{25}{4}-b^2$

$b^2=(\frac{25}{4}-b^2)(b^2-1)$

Let $b^2=t$

$t=(\frac{25}{4}-t)(t-1)$

$t=\frac{25}{4}t-t^2-\frac{25}{4}+t$

$4t^2-25t+25=0$

$4t^2-20t-5t+25=0$

$4t(t-5)-5(t-5)=0$

$t=\frac{5}{4},t=5$

$b^2=\frac{5}{4},5$

Substituting value of $b^2$ in (1)

we get $a^2=20,80$ for $b^2$ $=5,\frac{3}{4}$ respectively

Thus equation of ellipse is

$\frac{x^2}{20}+\frac{y^2}{5}=1$ and

$\frac{x^2}{80}+\frac{y^2}{5/4}=1$