Question

If the latus rectum of an ellipse is equal to half of the minor axis, then what is its eccentricity?
$\left( a \right)\dfrac{2}{\sqrt{3}}$
$\left( b \right)\dfrac{1}{\sqrt{3}}$
$\left( c \right)\dfrac{\sqrt{3}}{2}$
$\left( d \right)\dfrac{\sqrt{3}}{1}$

Answer 1

The eccentricity of the ellipse is given as $e=\dfrac{c}{a},$ so we will have to find the focal length (c). Firstly we have that latus rectum is the same as half the minor axis, i.e. $\dfrac{2{{b}^{2}}}{a}=b$ and we will solve this.

Complete step-by-step answer:
We get the relation between b and a as a = 2b which we will use it to find c given as $c=\sqrt{{{a}^{2}}-{{b}^{2}}}.$ Once, we have c, we will find the value of eccentricity (e).
We will consider that the equation of the ellipse is given as
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
So, we have the length of the major axis as $2\times a=2a.$
And the length of the minor axis as $2\times b=2b.$
Now, we are given that our ellipse has its latus rectum equal to the half of the minor axis.
We have the latus rectum of the ellipse given as
$\text{Latus Rectum}=\dfrac{2{{b}^{2}}}{a}$
So, as latus rectum is equal to the half of the minor axis,
$\Rightarrow \dfrac{2{{b}^{2}}}{a}=\dfrac{2b}{a}$
Simplifying further, we get,
$\Rightarrow 2{{b}^{2}}=ba$
Cancelling the like terms, we get,
$\Rightarrow 2b=a$
Now, the eccentricity of the latus rectum is given as $e=\dfrac{c}{a}$ where c is the focal length, a is the length of the semi-major axis.
So, to find e, we have to find the value of c first. We know that the focal length c is given as,
$c=\sqrt{{{a}^{2}}-{{b}^{2}}}$
As a = 2b. So,
$c=\sqrt{{{\left( 2b \right)}^{2}}-{{b}^{2}}}$
$\Rightarrow c=\sqrt{4{{b}^{2}}-{{b}^{2}}}$
Simplifying further, we get,
$c=\sqrt{3{{b}^{2}}}$
$\Rightarrow c=b\sqrt{3}$
As we get c as $b\sqrt{3},$ now we will find the value of eccentricity.
$e=\dfrac{c}{a}$
Here, $c=b\sqrt{3}$ and a = 2b. So, we get,
$\Rightarrow e=\dfrac{b\sqrt{3}}{2b}$
Cancelling the like term, we get,
$\Rightarrow e=\dfrac{\sqrt{3}}{2}$

So, the correct answer is “Option (c)”.

Note: Students make a simple mistake like $\sqrt{4{{b}^{2}}-{{b}^{2}}}=\sqrt{4}$ this happens. So be careful while subtracting as the variables are not eliminated when their coefficients are not the same. Also, remember that $2{{b}^{2}}=ba$ gives us 2b = a as b gets cancelled from both the sides. This happened because b is never zero. So, we can divide both sides by b and get 2b = a.