Let x^2+y^2=16 be the equation of the auxilia | vedclass

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(A) The auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$. $	herefore a^2=16 \Rightarrow a=4$. The hyperbola passe

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(A) The auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$. $	herefore a^2=16 \Rightarrow a=4$. The hyperbola passes through $(4 \sqrt{2}, 3)$,so: $\frac{(4 \sqrt{2})^2}{16} - \frac{3^2}{b^2} = 1$ $\Rightarrow \frac{32}{16} - \frac{9}{b^2} = 1$ $\Rightarrow 2 - \frac{9}{b^2} = 1$ $\Rightarrow \frac{9}{b^2} = 1$ $\Rightarrow b^2 = 9$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. $e = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

Let $x^2+y^2=16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and let $(4 \sqrt{2}, 3)$ be a point on the hyperbola. Then,the eccentricity of the hyperbola is

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