Question

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

• A. $\frac{x^2}{16}+\frac{y^2}{20}=1$
• B. $\frac{x^2}{20}+\frac{y^2}{9}=1$
• C. $\frac{x^2}{16}+\frac{y^2}{5}=1$
• D. $\frac{x^2}{20}+\frac{y^2}{5}=1$

Answer 1

The correct option is D $\frac{x^2}{20}+\frac{y^2}{5}=1$

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It passes through $(4,-1)$ $\therefore \frac{16}{a^2}+\frac{1}{b^2}=1$ or $a^2+16b^2=a^2{b}^2$
The given line $x+4y-10=0$ is $=-\frac{1}{4}x+\frac{5}{2}$ $\therefore m=-\frac{1}{4},c=\frac{5}{2}.$
The condition of tangency is $c^2=a^2{m}^2+b^2$ or $\frac{25}{4}=\frac{a^2}{16}+b^2=\frac{a^2+16b^2}{16}=\frac{a^2{b}^2}{16},$
$\therefore a^2{b}^2=100$ or $ab=10$ or $b=\frac{10}{a}$
Putting in (1), we get $\frac{16}{a^2}+\frac{1}{b^2}=1$ or $\frac{16}{a^2}+\frac{a^2}{100}=1$ or $a^4-100a^2+1600=0$ or $(a^2-80)(a^2-20)=0\therefore a^2=80,20$
Hence $b^2=\frac{100}{a^2}=\frac{100}{80}$ or $\frac{100}{20}=\frac{5}{4}$ or $5$
Hence the equations of the ellipse are $\frac{x^2}{20}+\frac{y^2}{5}=1$