Statement I: The eccentricity of the hyperbol | vedclass

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(A) Given equation: $9x^2-16y^2-72x+96y-144=0$Rearranging terms: $9(x^2-8x)-16(y^2-6y)=144$Completing the square: $9(x^2-8x+16)-16(y^2-6y+9)=144+144-1

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Statement $I$ is true,Statement $II$ is true; Statement $II$ is the correct explanation for Statement $I$.

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(A) Given equation: $9x^2-16y^2-72x+96y-144=0$Rearranging terms: $9(x^2-8x)-16(y^2-6y)=144$Completing the square: $9(x^2-8x+16)-16(y^2-6y+9)=144+144-144$$9(x-4)^2-16(y-3)^2=144$Dividing by $144$: $\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1$Comparing with $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$,we get $a^2=16$ and $b^2=9$.The eccentricity $e$ is given by $\sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.Thus,Statement $I$ is true and Statement $II$ is the correct formula used to derive it.

Statement $I$: The eccentricity of the hyperbola $9x^2-16y^2-72x+96y-144=0$ is $5/4$.
Statement $II$: The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sqrt{1+\frac{b^2}{a^2}}$.

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Statement $I$: The eccentricity of the hyperbola $9x^2-16y^2-72x+96y-144=0$ is $5/4$.Statement $II$: The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sqrt{1+\frac{b^2}{a^2}}$.