Question

If $3x+4y=12\sqrt{2}$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{9}=1$ for some $a\in R$, then the distance between the foci of the ellipse is :

• A. $2\sqrt{5}$
• B. $2\sqrt{7}$
• C. $2\sqrt{2}$
• D. $4$

Answer 1

The correct option is B $2\sqrt{7}$

$3x+4y=12\sqrt{2}$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{9}=1$
Equation of tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{9}=1$ is
$y=mx\pm \sqrt{a^2{m}^2+9}$

Now, $3x+4y=12\sqrt{2}$
$\Rightarrow y=-\frac{3}{4}x+3\sqrt{2}$
$\Rightarrow m=-\frac{3}{4}$ and $\sqrt{a^2{m}^2+9}=\pm 3\sqrt{2}$
$\Rightarrow a^2{(-\frac{3}{4})}^2+9=18$
$\Rightarrow a^2\times \frac{9}{16}=9$
$\Rightarrow a^2=16$

$c=\sqrt{a^2-b^2}=\sqrt{7}$
$\therefore$ Distance between the foci is $2c=2\sqrt{7}$