The eccentricity of the hyperbola $16 x^{2}-3 y^{2}-32 x-12 y-44=0$ is
The eccentricity of the hyperbola $16 x^{2}-3 y^{2}-32 x-12 y-44=0$ is
- $\sqrt{\frac{19}{3}}$
- $\sqrt{\frac{13}{19}}$
- $\frac{\sqrt{19}}{3}$
- $\frac{13}{\sqrt{19}}$
Solution
Equation: $16 x^{2}-3 y^{2}-32 x-12 y-444=0$
$16\left(x^{2}-2 x\right)-3\left(y^{2}-4 y\right)=44$
$\begin{array}{l}
16(x+1)^{2}-3(y-2)^{2}=48 \\
\frac{(x+1)^{2}}{3}-\frac{(y-2)^{2}}{16}=1
\end{array}$
Centre $(1,-2)$ bength of ransuene axis $=2 a$
$\begin{aligned}
\text { Length y con jugateaxis } &=2 b=2 \sqrt{3} \\
e &=\sqrt{\frac{1+b^{2}}{a^{2}}}=\sqrt{1+\left(\frac{4}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{1+16}{3}} \\
e &=\sqrt{\frac{19}{3}}
\end{aligned}$
Asked in: MHT CET 2020 (14 Oct Shift 1)
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