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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions: Vertices: $(0,\pm 3)$, Foci: $(0,\pm 5)$

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Given Vertices: $(0,\pm 3)$, Foci: $(0,\pm 5) \rightarrow a = 3, c = 5$
Since $a^2+b^2 = c^2 \rightarrow b = \sqrt{25-9} = \sqrt{16} = 4$
Therefore, the equation of the hyperbola is as follows: $\;\large\frac{y^2}{9}$$-\large\frac{x^2}{16}$$=1$
answered Apr 8, 2014 by balaji.thirumalai
 

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