Statement I The eccentricity of the hyperbola $9 x^2-16 y^2-72 x+96 y-144=0$ is $5 / 4$. Statement II The…
Statement I The eccentricity of the hyperbola $9 x^2-16 y^2-72 x+96 y-144=0$ is $5 / 4$.
Statement II The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sqrt{1+\frac{b^2}{a^2}}$
Statement I is true, Statement II is true; Statement II is correct explanation for Statement I.
Both statements are true and Statement II is not the correct explanation of Statement I.
Statement I is false; Statement II is true.
Statement I is true; Statement II is false.
Solution
$9 x^2-16 y^2-72 x+96 y-144=0$
$\begin{aligned} & \Rightarrow 9\left(x^2-8 x\right)-16\left(y^2-6 y\right)-144=0 \\ & \Rightarrow 9\left(x^2-8 x+16\right)-16\left(y^2-6 y+9\right)-144\end{aligned}$ $-144+144=0$
$\begin{aligned} & \Rightarrow \quad 9(x-4)^2-16(y-3)^2=144 \\ & \Rightarrow \quad \frac{(x-4)^2}{4^2}-\frac{(y-3)^2}{3^2}=1\end{aligned}$
Here, $a=4, b=3$
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
$\therefore$ Staement I and statement II are true and statement II is a correct explanation for Statement I.