If 3 x+4 y=12 is a tangent
Question:

If $3 x+4 y=12 \sqrt{2}$ 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:

1. (1) $2 \sqrt{7}$

2. (2) 4

3. (3) $2 \sqrt{5}$

4. (4) $2 \sqrt{2}$

Correct Option: 1

Solution:

$3 x+4 y=12 \sqrt{2}$

$\Rightarrow \quad 4 y=-3 x+12 \sqrt{2}$

$\Rightarrow \quad y=-\frac{3}{4} x+3 \sqrt{2}$

Now, condition of tangency, $c^{2}=a^{2} m^{2}+b^{2}$

$\therefore \quad 18=a^{2} \cdot \frac{9}{16}+9 \quad \Rightarrow \quad a^{2} \cdot \frac{9}{16}=9$

$\Rightarrow a^{2}=16 \Rightarrow a=4$

Eccentricity $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

$\therefore \quad a e=\frac{\sqrt{7}}{4} \cdot 4=\sqrt{7}$

$\therefore \quad$ Focus are $(\pm \sqrt{7}, 0)$

$\therefore \quad$ Distance between foci of ellipse $=2 \sqrt{7}$