Question

If the distance between foci of an ellipse is equal to the length of the latus rectum, then the eccentricity is,
A) $\dfrac{1}{4}\left( {\sqrt 5 - 1} \right)$
B) $\dfrac{1}{2}\left( {\sqrt 5 + 1} \right)$
C) $\dfrac{1}{2}\left( {\sqrt 5 - 1} \right)$
D) $\dfrac{1}{4}\left( {\sqrt 5 + 1} \right)$

Answer 1

To solve this question, what we will use the formula of latus rectum and foci and we will put them in equals as per the condition i.e. the distance between foci of an ellipse is equal to the length of the latus rectum and then, we will solve to get the value of eccentricity of an ellipse which equals to e.

Complete step-by-step solution:
The diagram for the question is,

An ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant. And, its equation is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.
The given figure is of an ellipse. It has two foci ${F_1}$ and ${F_2}$ which are equidistant from the center point and is equal to 2ae where e denotes eccentricity of the ellipse.
PQ and RS denote the latus rectum which is equal to $\dfrac{{2{b^2}}}{a}$.
Now, we know that the distance between foci is 2ae and the latus rectum is equal to $\dfrac{{2{b^2}}}{a}$. In question, it is given that the distance between foci is equal to that length of the latus rectum.
So, we can write the above condition as,
$\Rightarrow 2ae = \dfrac{{2{b^2}}}{a}$
Simplify the terms,
$\Rightarrow {a^2}e = {b^2}$
We know that ${b^2} = {a^2}\left( {1 - {e^2}} \right)$. Substitute in the above equation,
$\Rightarrow {a^2}e = {a^2}\left( {1 - {e^2}} \right)$
Cancel out ${a^2}$ from both sides,
$\Rightarrow e = 1 - {e^2}$
Move all terms on one side,
$\Rightarrow {e^2} + e - 1 = 0$
Using quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to evaluate the value of $e$, we get
$\Rightarrow e = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 1 \times - 1} }}{{2 \times 1}}$
Simplify the terms,
$\Rightarrow e = \dfrac{{ - 1 \pm \sqrt {1 + 4} }}{2}$
Add the term inside the bracket,
$\Rightarrow e = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$
Since the value of e cannot be negative. So,
$\therefore e = \dfrac{{ - 1 + \sqrt 5 }}{2}$

Hence, option (C) is correct.

Note: To solve this question, one must know what an ellipse is and its diagram as it helps in visualizing the question easily. Always remember that the general equation of an ellipse is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and $a{x^2} + bx + c = 0$, quadratic formula to evaluate the value of $x$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. The calculation should be done very accurately and carefully as it may affect the answer and substitute the values in such a way to make complex equations simpler.