JEE Main · 2016 · 03 Apr 2016 · Offline · Hyperbola

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The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal half of the distance between its foci, is:

(A) $\frac{4}{3}$
(B) $\frac{2}{3}$ ✓ Correct
(C) $3$
(D) $\frac{4}{3}$

Answer: (B)

Solution

For

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, a latus rectum of length 8 gives

\frac{2 b ^{2}}{a} = 8

, so

b^{2} = 4 a

. The conjugate axis being half the distance between the foci gives

2 b = \frac{1}{2} (2 a e)

, i.e.

2 b = a e

, so

4 b^{2} = a^{2} e^{2}

; with

b^{2} = 4 a

this is

16 a = a^{2} e^{2}

, hence

e^{2} = \frac{16}{a}

. Also

b^{2} = a^{2} (e^{2} - 1)

gives

e^{2} - 1 = \frac{4}{a}

. Equating,

\frac{16}{a} = 1 + \frac{4}{a}

, so

\frac{12}{a} = 1

and

a = 12

, making

e^{2} = \frac{16}{12} = \frac{4}{3}

and

e = \frac{2}{3}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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