QUESTION

# The length of the latus rectum of an ellipse is one third of the major axis, its eccentricity would beA. $\dfrac{2}{3}$B. $\sqrt{\dfrac{2}{3}}$C. $\dfrac{1}{\sqrt{3}}$D. $\dfrac{1}{\sqrt{2}}$

Hint: The length of the latus rectum of an ellipse is $\dfrac{2{{b}^{2}}}{a}$. Thus find the length which is ${{\dfrac{1}{3}}^{rd}}$ to the major axis. Substitute the obtained value in the formula of eccentricity of an ellipse.

Complete Step-by-Step solution:
We know the general form of an ellipse, $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
The latus rectum of an ellipse is the chord of the chord of the ellipse through its one focus and perpendicular to the major axis.
We know the latus rectum of ellipse $=\dfrac{2{{b}^{2}}}{a}$.
It is said that the length of the latus rectum of an ellipse is equal to ${{\dfrac{1}{3}}^{rd}}$ of the major axis.
We know the length of the major axis of ellipse = 2a.
$\therefore$ According to the question, $\dfrac{2{{b}^{2}}}{a}=\dfrac{1}{3}\left( 2a \right)$.
By simplifying the above expression, we get,
$\dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{1}{3}$
Now we need to find the eccentricity.
The eccentricity of an ellipse is the ratio of the distance from the center to the foci and the distance from the center of the vertices.
The equation of eccentricity is given by,
${{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$
Put, $\dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{1}{3}$ in the above expression.
\begin{align} & {{e}^{2}}=1-\dfrac{1}{3}=\dfrac{3-1}{3} \\ & {{e}^{2}}=\dfrac{2}{3} \\ & \therefore e=\sqrt{\dfrac{2}{3}} \\ \end{align}
Thus we got the eccentricity, e as $\sqrt{\dfrac{2}{3}}$.
$\therefore$ Option (b) is the correct answer.

Note: We know that ellipse is a closed shape structure in a two dimensional plane. Hence it covers a region in a 2D plane. So, this bounded region of the ellipse forms its area. The shape of the ellipse is different from that of the circle.
Area of ellipse = $\pi \times$ Major axis $\times$ Minor axis = $\pi ab$.