The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is half the distance between the foci is

(a) $\frac{3}{2}$
(b) $2$
(c) $\frac{3}{2}$
(d) None of these

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Alok Kumar Singh | Contributor-Level 10

7 months ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e n g t h o f t h e l a t u s r e c t u m o f t h e h y p e r b o l a \\ = \frac{2 b^{2}}{a} = 8 \Rightarrow b^{2} = 4 a \dots (i) \\ Distance b e t w e e n t h e f o c i = 2 a e \\ T r a n s v e r s e a x i s = 2 a \\ a n d C o n j u g a t e a x i s = 2 b \\ ∴ \frac{1}{2} (2 a e) = 2 b \Rightarrow a e = 2 b \Rightarrow b = \frac{a e}{2} \dots (i i) \\ \Rightarrow b^{2} = \frac{a^{2} e^{2}}{4} \Rightarrow 4 a = \frac{a^{2} e^{2}}{4} [F r o m e q n . (i)] \\ \Rightarrow 16 = a e^{2} ∴ a = \frac{16}{e^{2}} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 4 a = a^{2} (e^{2} - 1) \\ \Rightarrow \frac{4}{a} = e^{2} - 1 \Rightarrow \frac{4}{16 / e^{2}} = e^{2} - 1 \\ \Rightarrow \frac{e^{2}}{4} = e^{2} - 1 \Rightarrow e^{2} - \frac{e^{2}}{4} = 1 \Rightarrow \frac{3 e^{2}}{4} = 1 \\ \Rightarrow e^{2} = \frac{4}{3} ∴ e = \frac{2}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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