Question

# The area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$, is

Hint: We know that the ends of latus rectum of an ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ are $\left( \pm ae,\dfrac{{{b}^{2}}}{a} \right)$. We know that tangent of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ at $\left( {{x}_{1}},{{y}_{1}} \right)$ is $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$. We know that there will be four latus rectums for an ellipse. So, the area of a quadrilateral ellipse formed by all the latus rectums is equal to 4 times of the area of the triangle formed by a single latus rectum. So, we can find the area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$

We know that the eccentricity of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to $\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
Now we should find the eccentricity of $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$.
Let us compare $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ with $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$.
Then we get
\begin{align} & {{a}^{2}}=9.....(1) \\ & {{b}^{2}}=5.....(2) \\ \end{align}
Let us the eccentricity of $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to e.
$\Rightarrow e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}.....(3)$
Now, let us substitute equation (1) and equation (2) in equation (3), then we get
\begin{align} & \Rightarrow e=\sqrt{1-\dfrac{5}{9}} \\ & \Rightarrow e=\sqrt{\dfrac{9-5}{9}} \\ & \Rightarrow e=\sqrt{\dfrac{4}{9}} \\ & \Rightarrow e=\dfrac{2}{3}.....(4) \\ \end{align}

We know that the ends of latus rectum of an ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ are $\left( \pm ae,\dfrac{{{b}^{2}}}{a} \right)$.
Now we should find the value of $ae$.
From equation (1) and equation (4), we get
\begin{align} & \Rightarrow ae=\left( 3 \right)\left( \dfrac{2}{3} \right) \\ & \Rightarrow ae=2....(5) \\ \end{align}
Now we should find the value of $\dfrac{{{b}^{2}}}{a}$.
$\Rightarrow \dfrac{{{b}^{2}}}{a}=\dfrac{5}{3}..(6)$
From equation (5) and equation (6), we get that the ends of the latus rectum are $\left( \pm 2,\dfrac{5}{3} \right)$.
Now let us find the tangent equation of $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$ with $\left( 2,\dfrac{5}{3} \right)$.
We know that tangent of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ at $\left( {{x}_{1}},{{y}_{1}} \right)$ is $\dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$.
So, the equation of tangent equation of $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$ with $\left( 2,\dfrac{5}{3} \right)$ is
$\Rightarrow \dfrac{2x}{9}+\dfrac{\left( \dfrac{5}{3} \right)y}{5}=1$
$\Rightarrow \dfrac{2x}{9}+\dfrac{y}{3}=1......(7)$
From equation (7), it is clear that $\dfrac{2x}{9}+\dfrac{y}{3}=1$ is the tangent equation of $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$ at $\left( 2,\dfrac{5}{3} \right)$.
Now, we should find the area of the quadrilateral of $\dfrac{2x}{9}+\dfrac{y}{3}=1$ with respect to axes.
Now, we should find the point where $\dfrac{2x}{9}+\dfrac{y}{3}=1$ meet the x-axis.
So, let us substitute the value of y is equal to 0 in $\dfrac{2x}{9}+\dfrac{y}{3}=1$.
\begin{align} & \Rightarrow \dfrac{2x}{9}+\dfrac{y}{3}=1 \\ & \Rightarrow \dfrac{2x}{9}=1 \\ \end{align}
$\Rightarrow x=\dfrac{9}{2}.....(8)$
Now, we should find the point where $\dfrac{2x}{9}+\dfrac{y}{3}=1$ meet the y-axis.
So, let us substitute the value of x is equal to 0 in $\dfrac{2x}{9}+\dfrac{y}{3}=1$.
\begin{align} & \Rightarrow \dfrac{2x}{9}+\dfrac{y}{3}=1 \\ & \Rightarrow \dfrac{y}{3}=1 \\ \end{align}
$\Rightarrow y=3.....(9)$
So, it is clear that $\dfrac{2x}{9}+\dfrac{y}{3}=1$ meets x-axis at $A\left( \dfrac{9}{2},0 \right)$ and y-axis at $B\left( 0,3 \right)$.
We know that area of the triangle is equal to half of the product of OA and OB.
So, the area of the triangle OAB is equal to half of the product of OA and OB.
\begin{align} & \Rightarrow \text{Area of }\Delta \text{OAB=}\dfrac{1}{2}\left( OA \right)\left( OB \right) \\ & \Rightarrow \text{Area of }\Delta \text{OAB=}\dfrac{1}{2}\left( \dfrac{9}{2} \right)\left( 3 \right) \\ & \Rightarrow \text{Area of }\Delta \text{OAB=}\dfrac{27}{4}......(10) \\ \end{align}
From equation (1), it is clear that the area of the triangle formed by a latus rectum is equal to $\dfrac{27}{4}$.
We know that there will be four latus rectums for an ellipse.
So, the area of a quadrilateral ellipse formed by all the latus rectums is equal to 4 times of the area of the triangle formed by a single latus rectum.
Let us assume the area of the quadrilateral is equal to A.
\begin{align} & \Rightarrow A=4\left( \dfrac{27}{4} \right) \\ & \Rightarrow A=27....(11) \\ \end{align}
So, it is clear that the area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$ is 27.

Note: Students have a misconception that the ends of latus rectum of an ellipse are equal to $\left( \pm \dfrac{{{b}^{2}}}{a},ae \right)$. If this misconception is followed, then we get an incorrect equation of latus rectum. Then we will get the incorrect are of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1$. So, the misconception should be avoided.