The equation of the hyperbola with eccentricity 2 and foci at (±2,0) is (a) x 2 4 − y 2 4 = 1 (b) x 2 4 − y 2 5 = 1 (c) x 2 9 − y 2 4 = 1 (d) None of these

This is an objective Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t e = 3 2 a n d f o c i = ( ± a e , 0 ) = ( ± 2 , 0 ) ∴ a e = 2 ⇒ a * 3 2 = 2 ⇒ a = 4 3 W e k n o w t h a t b 2 = a 2 ( e 2 − 1 ) ⇒ b 2 = 1 6 9 ( 9 4 − 1 ) ⇒ b 2 = 1 6 9 * 5 4 ⇒ b 2 = 2 0 9 S o , t h e e q u a t i o n o f t h e h y p e r b o l a i s x 2 ( 4 / 3 ) 2 − y 2 2 0 / 9 = 1 ⇒ 9 x 2 1 6 − 9 y 2 2 0 = 1 ⇒ x 2 1 6 − y 2 2 0 = 1 9 ⇒ x 2 4 − y 2 5 = 4 9 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

The equation of the hyperbola with eccentricity $\sqrt[]{2}$ and foci at $(\pm 2, 0)$ is

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
(c) $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
(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:

^{$\begin{array}{l} G i v e n t h a t e = \frac{3}{2} \\ a n d f o c i = (\pm a e, 0) = (\pm 2, 0) \\ ∴ a e = 2 \Rightarrow a * \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) \Rightarrow b^{2} = \frac{16}{9} * \frac{5}{4} \\ \Rightarrow b^{2} = \frac{20}{9} \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{{(4 / 3)}^{2}} - \frac{y^{2}}{20 / 9} = 1 \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{20} = \frac{1}{9} \\ \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$}

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