Question

The eccentricity of the hyperbola whose length of the latus rectum is 8 and the length of its conjugate axis is equal to half of the distance between its foci is:
(A) $\dfrac{4}{{\sqrt 3 }}$
(B) $\dfrac{2}{{\sqrt 3 }}$
(C) $\sqrt 3 $
(D) $\dfrac{4}{3}$

Answer 1

Hint:We know latus rectum for hyperbola is given by $\dfrac{{2{b^2}}}{a}$ consider it as equation 1.They given length of its conjugate axis is equal to half of the distance between its foci now consider it as equation 2.Solving equations 1 and 2 we get value of $a$ and $b$, substitute in eccentricity formula to get the required answer.

Complete step-by-step answer:
For hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, the length of latus rectum is $\dfrac{{2{b^2}}}{a}$.
Given, Length of latus rectum of the hyperbola= $8$
$\therefore $$\dfrac{{2{b^2}}}{a} = 8$
On solving further, we get
$ \Rightarrow {b^2} = \dfrac{{8a}}{2}$
$ \Rightarrow {b^2} = 4a$ …. (1)
For hyperbola$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, the length of transverse axis is $2a$ while the length of conjugate axis is $2b$.
Also, the coordinates of its foci is $\left( { \pm ae,0} \right)$, hence the distance between its foci is $2ae$.
Given that-
Length of conjugate axis of hyperbola= Half of the distance between its foci
$2b = \dfrac{1}{2}\left( {2ae} \right)$
On solving further, we get
$ \Rightarrow b = \dfrac{{ae}}{2}$…. (2)
Substitute the value of $b$ from equation (2) to equation (1),we get
${\left( {\dfrac{{ae}}{2}} \right)^2} = 4a$
$ \Rightarrow \dfrac{{{a^2}{e^2}}}{4} = 4{a^{}}$
$ \Rightarrow {a^2}{e^2} = 16a$ …. (3)
But for hyperbola$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ we also know that,
 ${b^2} = {a^2}{e^2} - {a^2}$
Now, substitute the values of ${b^2}$ and ${a^2}{e^2}$ from equation (1) and (3), we get
$ \Rightarrow 4a = 16a - {a^2}$
$ \Rightarrow {a^2} - 12a = 0$
$ \Rightarrow a\left( {a - 12} \right) = 0$
$ \Rightarrow $$a = 12$
Now, Substitute $a = 12$ in equation (1)to find the value of $b$:
${b^2} = 4\left( {12} \right)$
${b^2} = 48$
$b = \sqrt {48} $
Now eccentricity for hyperbola is given by $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $
$e = \sqrt {1 + \dfrac{{{{\left( {\sqrt {48} } \right)}^2}}}{{{{\left( {12} \right)}^2}}}} $
$ \Rightarrow e = \sqrt {1 + \dfrac{{48}}{{144}}} $
$ \Rightarrow e = \sqrt {1 + \dfrac{1}{3}} $
$ \Rightarrow e = \sqrt {\dfrac{4}{3}} $
$ \Rightarrow e = \dfrac{2}{{\sqrt 3 }}$

So, the correct answer is “Option B”.

Note:The eccentricity of a conic section tells us how close it is to being in the shape of a circle. The eccentricity of a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is always greater than$1$ and can be calculated by the formula: $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $.While solving the problems related to hyperbola i.e., $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ , always remember the relation ${b^2} = {a^2}{e^2} - {a^2}$ ,which is an indirect form of the formula $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $.