Question

If the line $3x+4y=12\sqrt{2}$ is a tangent to the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{9}=1$ for some $a\in R$ , then the distance between the foci of the ellipse is:
$\begin{align}
  & A.\text{ }2\sqrt{7} \\
 & B.\text{ 4} \\
 & C.\text{ }2\sqrt{5} \\
 & D.\text{ }2\sqrt{2} \\
\end{align}$

Answer 1

Hint: In this question, we have to find the distance between the foci. Thus, we will use the general equation and the condition of the tangency to get the solution. First, we will get a new equation from the given equation in terms of x. After that, we will compare the general form of the equation and the new equation to get the slope and the intercept. Then, we will substitute these values in the condition of tangency ${{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}$ to get the value of a. Then, we will find the value of eccentricity where $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . in the last, we will find the foci and the distance between both the foci, to get the required solution for the problem.

Complete step-by-step solution:
According to the problem, we have to find the distance between the foci.
Thus, we will use the general form of equation and the condition of the tangency to get the solution
The equation given to us is $3x+4y=12\sqrt{2}$ --------- (1)
Now, we will first get a new equation for y in terms of x, that is we will subtract 3x on both sides in equation (1), we get
$\Rightarrow 3x+4y-3x=12\sqrt{2}-3x$
As we know, the same terms with opposite signs cancel out each other, we get
$\Rightarrow 4y=12\sqrt{2}-3x$
Now, we will divide 4 on both sides in the above equation, we get
$\Rightarrow \dfrac{4y}{4}=\dfrac{12\sqrt{2}-3x}{4}$
On splitting the denominator with respect to addition on the right hand side of the above equation, we get
$\Rightarrow y=\dfrac{12\sqrt{2}}{4}-\dfrac{3x}{4}$
On further solving the above equation, we get
$\Rightarrow y=-\dfrac{3x}{4}+3\sqrt{2}$ --------- (2)
Now, we know the general form of the equation is $y=mx+c$ , where m is the slope of the equation and c is the intercept. Thus, on comparing the general form of equation and equation (2), we get
$m=-\dfrac{3}{4}$ and $c=3\sqrt{2}$ ------ (3)
Now, we know that the general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the given ellipse equation is equal to $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{9}=1$ for some $a\in R$ , thus on comparing both the equations, we get $b=3$ and $a=a$ -------- (4)
Now, we know the condition of tangency is equal to ${{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}$ . Thus, on substituting equation (3) and (4) in the above formula, we get
$\Rightarrow {{\left( 3\sqrt{2} \right)}^{2}}={{a}^{2}}{{\left( \dfrac{-3}{4} \right)}^{2}}+{{\left( 3 \right)}^{2}}$
$\Rightarrow 18=\dfrac{9}{16}{{a}^{2}}+9$
Now, we will subtract 9 on both sides of the equation, we get
$\Rightarrow 18-9=\dfrac{9}{16}{{a}^{2}}+9-9$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 9=\dfrac{9}{16}{{a}^{2}}$
Now, we will multiply $\dfrac{16}{9}$ on both sides in the above equation, we get
$\Rightarrow 9\times \dfrac{16}{9}=\dfrac{9}{16}{{a}^{2}}\times \dfrac{16}{9}$
On further simplification, we get
$\Rightarrow 16={{a}^{2}}$
On taking the square root on both sides of the equation, we get
$\Rightarrow \sqrt{16}=\sqrt{{{a}^{2}}}$
$\Rightarrow \pm 4=a$ ----------- (5)
Now, we will find the eccentricity of the ellipse, where $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ . So, we will substitute the value of equation (4) and (5) in the above formula, we get
$\Rightarrow e=\sqrt{1-\dfrac{{{\left( 3 \right)}^{2}}}{{{\left( 4 \right)}^{2}}}}$
$\Rightarrow e=\sqrt{1-\dfrac{9}{16}}$
On taking the LCM on the denominator in the above equation, we get
$\Rightarrow e=\sqrt{\dfrac{16-9}{16}}=\sqrt{\dfrac{7}{16}}$
Now, on further simplification, we get
$\Rightarrow e=\pm \dfrac{\sqrt{7}}{4}$ ---------- (6)
Thus, we will find the $ae$ , by putting the value of equation (5) and (6), we get
$\Rightarrow ae=\left( \pm \dfrac{\sqrt{7}}{4} \right)\times \left( \pm 4 \right)$
$\Rightarrow ae=\pm \sqrt{7}$
Therefore, focus of the ellipse is equal to $\left( focus,0 \right)$ which is equal to $\left( \pm \sqrt{7},0 \right)$
Thus, the distance between the foci is equal to subtraction of the distance from the center to the vertex and the distance from the center to a co-vertex, that is $c=a-b$ , which is equal to
$\Rightarrow \sqrt{7}+\sqrt{7}$
$\Rightarrow 2\sqrt{7}$ which is the solution.
Therefore, if $3x+4y=12\sqrt{2}$ is a tangent to the ellipse

Note: While solving this problem, do step and step calculation to get the solution. Do not forget to mention the $\pm $ sign wherever it is necessary. Remember that for this problem, we have to find the value of ‘a’ and then solve further to get the accurate solution.