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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions: Foci: $(\pm 3\sqrt 5, 0)$, Latus Rectum is of length $8$.

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Given Foci: $(\pm 3\sqrt 5, 0) \rightarrow c = 3\sqrt5$.
Given Latus Rectum is of length $8 \rightarrow 8 = \large\frac{2b^2}{a}$$\rightarrow b^2 = 4a$
Since $a^2+b^2 = c^2 \rightarrow a^2+4a=3^2\sqrt5^2 \rightarrow a = 5$ (We discard $a=-9$ since $a$ is non-negative).
If $a = 5 \rightarrow b^2 = 4 \times 5 = 20$
Therefore, the equation of the hyperbola is as follows: $\;\large\frac{x^2}{25}$$-\large\frac{y^2}{20}$$=1$
answered Apr 8, 2014 by balaji.thirumalai
 

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