Question

An ellipse passes through the poitnt $(4,-1)$ and touches the line $x+4y-10=0$. if its axes concides with co-ordinates axes, then its equation is $(a>b)$

• A. $\frac{x^2}{16}+\frac{y^2}{15}=1$
• B. $\frac{x^2}{80}+\frac{y^2}{5}=1$
• C. $\frac{x^2}{20}+\frac{y^2}{5}=1$
• D. $\frac{x^2}{5}+\frac{y^2}{16}=1$

Answer 1

Answer: (C)

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

Solution:- (C) $\frac{x^2}{20}+\frac{y^2}{5}=1$

$x+4y=10$

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

Let equation of ellipse be-

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

As we know that, $y=mx+c$ is a tangent to ellipse if $c^2=a^2{m}^2+b^2$

From equation $\left(1\right)$, we get

$c=\frac{5}{2}$

$m=-\frac{1}{4}$

$\therefore \frac{25}{4}=a^2\left(\frac{1}{16}\right)+b^2$

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

As $(4,-1)$ lies on equation $\left(2\right)$.

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

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

$a^2=20$

$b^2=5$

$\therefore$ Equation of ellipse will be-

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

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