Question

An ellipse passes through the point $(4,-1)$ and its axes are along the axes of co-ordinates. If the line $x+4y-10=0$ is a tangent to it then its equation is

• A. $\frac{x^2}{100}+\frac{y^2}{5}=1$
• B. $\frac{x^2}{80}+\frac{y^2}{\frac{5}{4}}=1$
• C. $\frac{x^2}{20}+\frac{y^2}{5}=1$
• D. $Noneofthese$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{x^2}{80}+\frac{y^2}{\frac{5}{4}}=1$
C $\frac{x^2}{20}+\frac{y^2}{5}=1$

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It passes through $(4,-1)$
$\therefore$ $\frac{16}{a^2}+\frac{1}{b^2}=1$...(1)
$y=-\frac{1}{4}x+\frac{5}{2}$ is a tangent, then $c^2=a^2{m}^2+b^2$ or $\frac{25}{4}=a^2\cdot \frac{1}{16}+b^2...\left(2\right)$
Eliminating $b^2,$ we get $a^4-100a^2+1600=0$ or $(a^2-80)(a^2-20)=0\therefore a^2=80,20$

Corresponding values of $b^2$ are $\frac{5}{4}$ and $5$

Hence, the ellipse are as given in (b) and (c).