Question

If the eccentricity of an ellipse is $\dfrac{5}{8}$ and distance between foci is $10$, then the length of the latus rectum of the ellipse.

Answer 1

Hint: First we will use the formula for foci of an ellipse that is $2ae = f$, to find the value of $a$. Then we will have the value of both $e$ and $a$ and can hence find the value of $b$ by using the formula$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} $. On finding the value of both of $a$ and $b$, we can easily find the length of the latus rectum of the ellipse by using the formula $l = \dfrac{{2{b^2}}}{a}$, which is required for us to find.

Complete step by step answer:
We are given, the eccentricity of the ellipse is, $e = \dfrac{5}{8}$.
And, the distance between the foci is, $f = 10$.
So, we know, for an ellipse, $2ae = f$.
Substituting the known values in the above formula, we get,
$2.a.\dfrac{5}{8} = 10$
$ \Rightarrow a.\dfrac{{10}}{8} = 10$
Multiplying both sides of the equation by $\dfrac{8}{{10}}$, we get,
$ \Rightarrow a = 10.\dfrac{8}{{10}}$
$ \Rightarrow a = 8$
Therefore, from the given data, we get, $a = 8$.

Now, we know, the formula for eccentricity is,
$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} $
Squaring on both sides, we get,
$ \Rightarrow {e^2} = 1 - {\left( {\dfrac{b}{a}} \right)^2}$
Subtracting $1$ from both sides, gives us,
$ \Rightarrow {e^2} - 1 = - {\left( {\dfrac{b}{a}} \right)^2}$
Multiplying both sides by $ - 1$, we get,
$ \Rightarrow {\left( {\dfrac{b}{a}} \right)^2} = 1 - {e^2}$
$ \Rightarrow \dfrac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$
Multiplying both sides by ${a^2}$, we get,
$ \Rightarrow {b^2} = {a^2}(1 - {e^2}) - - - \left( 1 \right)$

Now, we know, the length of latus rectum is,
$l = \dfrac{{2{b^2}}}{a} - - - \left( 2 \right)$
Substituting $\left( 1 \right)$ in $\left( 2 \right)$, we get,
$ \Rightarrow l = \dfrac{{2.{a^2}\left( {1 - {e^2}} \right)}}{a}$
$ \Rightarrow l = 2.a\left( {1 - {e^2}} \right)$
Now, substituting $a = 8$ and $e = \dfrac{5}{8}$, in the above equation, we get,
$ \Rightarrow l = 2.8.\left[ {1 - {{\left( {\dfrac{5}{8}} \right)}^2}} \right]$
$ \Rightarrow l = 16.\left[ {1 - \dfrac{{25}}{{64}}} \right]$
Now, taking the LCM, we get,
$ \Rightarrow l = 16.\left[ {\dfrac{{39}}{{64}}} \right]$
$ \Rightarrow l = \dfrac{{39}}{4}$
$ \therefore l = 9\dfrac{3}{4}$

Therefore, the length of the latus rectum is $9\dfrac{3}{4}$.

Note: The eccentricity of a conic section differs from one to another. Like, for parabola, the eccentricity is always $1$. While, for an ellipse, the eccentricity is always less than $1$. On the other hand, for hyperbola, the eccentricity is always greater than $1$.