QUESTION

# Find the equation of an ellipse whose latus rectum is 8 and eccentricity is $\dfrac{1}{3}$.

Hint: Here, we are given the length of the latus rectum, $\dfrac{2{{b}^{2}}}{a}=8$ and eccentricity $e=\dfrac{1}{3}$. First we have to find the value of $a$ by substituting the formula ${{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$. Then after getting $a$ find the value of $b$ from latus rectum. Substitute $a$ and $b$ in the standard equation of ellipse to get the required equation.

We know that the equation of an ellipse whose centre (0, 0) and major axis parallel to x-axis is:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$where a > b and ${{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$, e is the eccentricity. It has a latus rectum where:
The length of the latus rectum = $\dfrac{2{{b}^{2}}}{a}$.
Here it is given that the length of the latus rectum, $\dfrac{2{{b}^{2}}}{a}=8$
Eccentricity, $e=\dfrac{1}{3}$
Now, we can write:
$\dfrac{2{{b}^{2}}}{a}=8$ where,
${{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$
Now by substituting the value of ${{b}^{2}}$ we get:
$\dfrac{2{{a}^{2}}\left( 1-{{e}^{2}} \right)}{a}=8$
By cancellation and substituting $e=\dfrac{1}{3}$, we get:
\begin{align} & 2a\left( 1-{{\left( \dfrac{1}{3} \right)}^{2}} \right)=8 \\ & 2a\left( 1-\dfrac{1}{9} \right)=8 \\ \end{align}
Next, by taking the LCM we get:

\begin{align} & 2a\left( \dfrac{9-1}{9} \right)=8 \\ & 2a\left( \dfrac{8}{9} \right)=8 \\ \end{align}
Next, by cross multiplication we obtain:
$a\left( \dfrac{8}{9} \right)=\dfrac{8}{2}$
Next, by the cancellation of 8 by 2 we get:
$\dfrac{8}{9}a=4$
Now, again by cross multiplication we obtain:
$a=\dfrac{4\times 9}{8}$
Next, by cancellation we obtain:
$a=\dfrac{9}{2}$
Now, by taking square on both the sides we get:
\begin{align} & {{a}^{2}}={{\left( \dfrac{9}{2} \right)}^{2}} \\ & {{a}^{2}}=\dfrac{81}{4} \\ \end{align}
Now, find the value of b using the formula:
$\dfrac{2{{b}^{2}}}{a}=8$
By cross multiplication we get:
$2{{b}^{2}}=8a$
Now, by substituting the value of $a=\dfrac{9}{2}$ we obtain:
$2{{b}^{2}}=8\times \dfrac{9}{2}$
Next, by cancellation we get:
\begin{align} & 2{{b}^{2}}=4\times 9 \\ & 2{{b}^{2}}=36 \\ \end{align}
Next, again by cross multiplication we obtain:
${{b}^{2}}=\dfrac{36}{2}$
Now again by cancellation we will get:
${{b}^{2}}=18$
Now, substitute the values of ${{a}^{2}}$ and ${{b}^{2}}$in the standard equation of ellipse:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Hence we will get:
$\dfrac{{{x}^{2}}}{\dfrac{81}{4}}+\dfrac{{{y}^{2}}}{18}=1$
We know that, $\dfrac{a}{\dfrac{b}{c}}=\dfrac{ac}{b}$
Therefore our equation becomes:
$\dfrac{4{{x}^{2}}}{81}+\dfrac{{{y}^{2}}}{18}=1$
Hence, $\dfrac{4{{x}^{2}}}{81}+\dfrac{{{y}^{2}}}{18}=1$ is the required equation of an ellipse.

Note: For an ellipse always a > b. If you are getting a < b then it won’t form the equation of an ellipse. So, after getting the values of a and b just check whether a > b or not.