Question

The equation of the ellipse whose axes are coincident with the co-ordinates axes 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
$y=mx\pm \sqrt{a^2{m}^2+b^2}$ (i)
since $3x-2y-20=0$ or $y=\frac{3}{2}x-10$ is tangent to the ellipse.
Comparing with Equation (i), $m=\frac{3}{2}$ and $a^2{m}^2+b^2=100$
$\Rightarrow a^2\times \frac{9}{4}+b^2=100$
$\Rightarrow 9a^2+4b^2=400$ (ii)
Similarly, since $x+6y-20=0$, i.e., $y=-\frac{1}{6}x+\frac{10}{3}$
is tangent to the ellipse, therefore comparing with Equation (i),
$m=-\frac{1}{6}$ and $a^2{m}^2+b^2=\frac{100}{9}$
$\Rightarrow \frac{a^2}{36}+b^2=\frac{100}{9}$

$\Rightarrow a^2+36b^2=400$ (iii)

Solving Equations (ii) and (iii), we get $a^2=40$ and $b^2=10$
Therefore, the required equation of the ellipse is
$\frac{x^2}{40}+\frac{y^2}{10}=1$