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

$2\sqrt{7}$

Explanation for correct option:

Step:1 Calculate the tangent to the ellipse and then compare it with the given equation.

Here, the given equation of the ellipse is $(\frac{x^2}{a^2}+\frac{y^2}{9})=1$.

The tangent to the ellipse is $y=mx+\sqrt{(a^2{m}^2+9)}$.

Given, the tangent to the ellipse is:

$\begin{array}{rcl}3x+4y& =& 12\sqrt{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& -\frac{3}{4}x+3\sqrt{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}-\frac{3x}{4}& =& mx\end{array}$

$\Rightarrow$ $\begin{array}{rcl}m& =& -\frac{3}{4}\end{array}$

Step:2 Calculate the value of $a$ and $e$

Again,

$\sqrt{(a^2{m}^2+9)}=3\sqrt{2}$

$\Rightarrow$$(a^2{m}^2+9)=18$

$\Rightarrow$ $m{}^2=a^{2}9$

$\Rightarrow$ $a^2{(-\frac{3}{4})}^2=9$

$\Rightarrow$ $a^2=4^2$

$\Rightarrow$ $a=4$

We know,

$\begin{array}{rcl}e& =& \sqrt{1-\frac{b^2}{a^2}}\\ & =& \sqrt{1-\frac{9}{16}}\\ & =& \frac{\sqrt{7}}{4}\end{array}$

Step:3 Calculate the distance between the foci of the given ellipse

$\therefore$the distance between foci $=\mathbf{2}\mathit{a}\mathit{e}$

$$\begin{gathered} =2\times 4\times \left(\frac{\sqrt{7}}{4}\right) \\ =\mathbf{2}\sqrt{\mathbf{7}} \end{gathered}$$

Hence, Option (B) is the correct answer.