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 $\left( 2 \right)$,

& a+ae=6 \\\ & \Rightarrow a(1+e)=6\,...............\left( 4 \right) \\\ \end{aligned}$$ Eccentricity: $$e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\,...............\left( 5 \right)$$ Now we can substitute e from equation (5) in equation (4), $$\begin{aligned} & \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{aligned}$$ So we can use equation (3) and substitute for $${{b}^{2}}$$, $$\begin{aligned} & \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{aligned}$$ Now squaring both sides, $$\begin{aligned} & \Rightarrow {{a}^{2}}-2a={{(6-a)}^{2}} \\\ & \Rightarrow {{a}^{2}}-2a=36+{{a}^{2}}-12a \\\ \end{aligned}$$ Cancelling similar terms from both sides we get, $$\begin{aligned} & \Rightarrow 10a=36 \\\ & \Rightarrow a=\dfrac{36}{10}=\dfrac{18}{5}\,...............\left( 6 \right) \\\ \end{aligned}$$ Now substituting the value of a from equation (6) in equation (4), $$\begin{aligned} & \dfrac{18}{5}(1+e)=6 \\\ & \Rightarrow (1+e)=\dfrac{5}{3} \\\ & \Rightarrow e=\dfrac{5}{3}-1 \\\ & \Rightarrow e=\dfrac{2}{3} \\\ \end{aligned}$$ Hence, the answer is option (a). Note: Remembering all the formulas is the key here and if we get one formula wrong we will surely commit a mistake. Substituting e is an important step. We may in hurry can tick mark option (b) or we may forget to subtract $$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.