Find the equation of the hyperbola with eccentricity 2 and foci at (±2,0) .

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Given that e=32 and foci=(±2,0)We know that foci=(±ae,0)∴ ae=2 ⇒a*32=2 ⇒a=43⇒ a2=169We know that b2=a2(e2−1)⇒ b2=169(94−1)=169*54=209So, the equation of the hyperbola is x216/9−y220/9=1⇒ 9x216−9y220=1 ⇒x24−y25=49Hence, the required equation is x24−y25=49.

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

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Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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(1) $\bar{E} : \frac{6!}{2!} = 360$

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