Question

The equation of the ellipse, whose axes are coincident with the co-ordinates axis and which touches the straight lines $3x-2y-20=0$ and $x+6y-20=0,$ is

• A. $\frac{x^2}{40}+\frac{y^2}{10}=1$
• B. $\frac{x^2}{5}+\frac{y^2}{8}=1$
• C. $\frac{x^2}{10}+\frac{y^2}{40}=1$
• D. $\frac{x^2}{40}+\frac{y^2}{30}=1$

Answer 1

The correct option is A $\frac{x^2}{40}+\frac{y^2}{10}=1$

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We know that the general equation of the tangent to the ellipse is
$y=mx\pm \sqrt{a^2{m}^2+b^2}\dots \left(i\right)$

Since $3x-2y-20=0$ or $y=\frac{3}{2}x-10$ is tangent to the ellipse.
$\therefore m=\frac{3}{2}$ and $a^2{m}^2+b^2=100$
$\Rightarrow a^2\times \frac{9}{4}+b^2=100$
$\Rightarrow 9a^2+4b^2=400\dots \left(ii\right)$

Also, $x+6y-20=0,$ or $y=-\frac{1}{6}x+\frac{10}{3}$ is tangent to the ellipse.
$\therefore m=-\frac{1}{6}$ and $a^2{m}^2+b^2=\frac{100}{9}$
$\Rightarrow \frac{a^2}{36}+b^2=\frac{100}{9}$
$\Rightarrow a^2+36b^2=400\dots \left(iii\right)$

Solving equations $\left(ii\right)$ and $\left(ii\right),$ we get
$a^2=40$ and $b^2=10$

Therefore, the required equation of the ellipse is
$\frac{x^2}{40}+\frac{y^2}{10}=1$