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# The foci of a hyperbola coincide with the foci of the ellipse $\large\frac{x^2}{25}+\frac{y^2}{9}$$=1.Find the equation of the hyperbola if its ecentricity is 3? \begin{array}{1 1}(a)\;\large\frac{x^2}{1}-\frac{y^2}{8}=\frac{16}{9}\\(b)\;\large\frac{x^2}{3}-\frac{y^2}{4}=\frac{15}{9}\\(c)\;x^2-y^2=9\\(d)\;\text{None of these}\end{array} Can you answer this question? ## 1 Answer 0 votes \large\frac{x^2}{25}+\frac{y^2}{9}$$=1$
$e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{9}{25}}$
$\Rightarrow \large\frac{4}{5}$
Foci of ellipse are $(\pm ae,0)$ i.e $(\pm 4,0)$
Let 'e' be the ecentricity of the required hyperbola and its equation be