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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: $(\pm 4 0)$, Latus Rectum is of length $12$.

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Given Foci: $(\pm 4, 0) \rightarrow c = 4$.
Given Latus Rectum is of length $12 \rightarrow 112 = \large\frac{2b^2}{a}$$\rightarrow b^2 = 6a$
Since $a^2+b^2 = c^2 \rightarrow a^2+6a=4\rightarrow a = 2$ (We discard $a=-8$ since $a$ is non-negative).
If $a = 2 \rightarrow b^2 = 6 \times 2 = 12$
Therefore, the equation of the hyperbola is as follows: $\;\large\frac{x^2}{4}$$-\large\frac{y^2}{12}$$=1$
answered Apr 8, 2014 by balaji.thirumalai
 

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