Question

If the line x – 2y = 12 is the tangent to the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ at the point $\left( 3,\dfrac{-9}{2} \right),$ then the length of the latus rectum of the ellipse is:
$\left( a \right)9$
$\left( b \right)8\sqrt{3}$
$\left( c \right)12\sqrt{2}$
$\left( d \right)5$

Answer 1

We are given the ellipse as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1.$ We will find the equation of the tangent using the formula $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$ at the point $\left( {{x}_{1}},{{y}_{1}} \right).$ Then we have x – 2y = 12 as the tangent. So, we will compare both the equations of the tangent, then we will find ${{a}^{2}}$ as 36 and ${{b}^{2}}$ as 27. So, at last, we will use the length of the latus rectus $\dfrac{2{{b}^{2}}}{a}$ to find the required answer.

Complete step-by-step answer:
We are given that we have the ellipse as
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
We will find the tangent of the ellipse. We know that for any ellipse, the equation of the tangent at $\left( {{x}_{1}},{{y}_{1}} \right)$ is given as
$\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$
So, at the point $\left( 3,\dfrac{-9}{2} \right)$ the tangent for the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ will be $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$ where ${{x}_{1}}=3$ and ${{y}_{1}}=\dfrac{-9}{2}.$
So, we get the equation of the tangent of the ellipse as
$\text{Equation of tangent}=\dfrac{3x}{{{a}^{2}}}-\dfrac{9y}{2{{b}^{2}}}=1.....\left( i \right)$
Also, we are given that for the same ellipse, $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ the tangent is
$x=2y=12.....\left( ii \right)$
So, those two tangents must be equal. So, we will compare those two equations. To compare, first, we will divide equation (ii) by 12 so that the right side is 1. Hence we get the equation of the tangent for the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{x}{12}-\dfrac{y}{6}=1.$
Now, we will compare $\dfrac{x}{12}-\dfrac{y}{6}=1$ and $\dfrac{3x}{{{a}^{2}}}-\dfrac{9y}{2{{b}^{2}}}=1.$ So, we get, $\dfrac{1}{12}=\dfrac{3}{{{a}^{2}}}$ and $\dfrac{-1}{6}=\dfrac{-9}{2{{b}^{2}}}.$
Now, after solving, we get,
$\Rightarrow \dfrac{1}{12}=\dfrac{3}{{{a}^{2}}}$
$\Rightarrow {{a}^{2}}=3\times 12$
Simplifying further we get,
${{a}^{2}}=36$
$\Rightarrow a=\pm 6$
Similarly, solving $\dfrac{-1}{6}=\dfrac{-9}{2{{b}^{2}}}$ gives us ${{b}^{2}}=\dfrac{-9\times -6}{2}.$
Simplifying, we get,
${{b}^{2}}=27$
$\Rightarrow b=\pm 3\sqrt{3}$
Now, we have to find the length of the latus rectum. We know that the length of the latus rectum of the ellipse is given as $\dfrac{2{{b}^{2}}}{a}.$
As a = 6 and ${{b}^{2}}=27,$ we get,
$\text{Length of the latus rectum}=\dfrac{2\times 27}{6}$
Simplifying, we get, the length of the latus rectum as 9.

So, the correct answer is “Option A”.

Note: While finding the latus rectum $\dfrac{2{{b}^{2}}}{a}$ we use a as 6 and do not use – 6 because the length of the latus rectum is the length which cannot be negative if we choose a = – 6 which gives us $\dfrac{2{{b}^{2}}}{a}=9$ which is not possible. So, we choose a as 6.