Question

If the length of latus rectum of an ellipse is 4 units and the distance between a focus and its farthest vertex on the major axis is 6 units, then its eccentricity is
(a) \[\dfrac{2}{3}\]
(b) \[\dfrac{1}{3}\]
(c) \[\dfrac{1}{5}\]
(d) \[\dfrac{1}{12}\]

Answer 1

Hint: As length of latus rectum is given 4 units we can get b in terms of a. Formula for length of latus rectum is \[\dfrac{2{{b}^{2}}}{a}\]which is same for both type of ellipse that is for a>b and b>a and similarly eccentricity is same for both ellipse.

Complete step by step answer:
Before proceeding with the question, we must know the definition of the ellipse. An ellipse is the set of all points in a plane, the sum of whose distances from two distinct fixed points (foci) is constant. Standard equation of ellipse is \[\dfrac{{{x}^{2}}}{{{a}^{2}}}\text{+}\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]\[(a>b)\]and \[\dfrac{{{x}^{2}}}{{{b}^{2}}}\text{+}\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] \[(b>a)\].
Here, a = length of semi-major axis and b = length of semi-minor axis. Major axis is horizontal when \[(a>b)\]. Major axis is vertical when \[(b>a)\]. The eccentricity of an ellipse is a measure of how nearly circular the ellipse is. Mathematically,
Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}............(1)\]
Distance between one focus and farthest vertex on major axis \[=a+ae................(2)\]
In this question, it is mentioned that the length of the latus rectum is 4 units and distance between a focus and its farthest vertex on the major axis is 6 units.

So from \[\left( 1 \right)\],
\[\dfrac{2{{b}^{2}}}{a}=4\] which implies \[{{b}^{2}}=2a...............\left( 3 \right)\]

So from \[\left( 2 \right)\],
\[\begin{align}
  & a+ae=6 \\
 & \Rightarrow a(1+e)=6\,...............\left( 4 \right) \\
\end{align}\]

Eccentricity: \[e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\,...............\left( 5 \right)\]

Now we can substitute e from equation (5) in equation (4),
\[\begin{align}
  & \Rightarrow \text{a(1+}\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\,)=6 \\
 & \Rightarrow \text{a(1+}\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}\,)=6 \\
\end{align}\]

So we can use equation (3) and substitute for \[{{b}^{2}}\],
\[\begin{align}
  & \Rightarrow \text{a(1+}\dfrac{\sqrt{{{a}^{2}}-2a}}{a}\,)=6 \\
 & \Rightarrow a\text{+}\sqrt{{{a}^{2}}-2a}\,=6 \\
 & \Rightarrow \sqrt{{{a}^{2}}-2a}\,=6-a \\
\end{align}\]

Now squaring both sides,
\[\begin{align}
  & \Rightarrow {{a}^{2}}-2a={{(6-a)}^{2}} \\
 & \Rightarrow {{a}^{2}}-2a=36+{{a}^{2}}-12a \\
\end{align}\]

Cancelling similar terms from both sides we get,
\[\begin{align}
  & \Rightarrow 10a=36 \\
 & \Rightarrow a=\dfrac{36}{10}=\dfrac{18}{5}\,...............\left( 6 \right) \\
\end{align}\]

Now substituting the value of a from equation (6) in equation (4),
\[\begin{align}
  & \dfrac{18}{5}(1+e)=6 \\
 & \Rightarrow (1+e)=\dfrac{5}{3} \\
 & \Rightarrow e=\dfrac{5}{3}-1 \\
 & \Rightarrow e=\dfrac{2}{3} \\
\end{align}\]

Hence, the answer is option (a).

Note: Remembering all the formulas is the key here and if we get one formula wrong we will surely commit a mistake. Substituting e is an important step. We may in hurry can tick mark option (b) or we may forget to subtract \[1\] from \[\dfrac{5}{3}\]and hence can think \[\dfrac{5}{3}\]as the answer which will not match with any options and we will end up guessing.