MHT CET · Maths · Hyperbola
The eccentricity of the hyperbola \(16 x^{2}-3 y^{2}-32 x-12 y-44=0\) is
- A \(\sqrt{\frac{19}{3}}\)
- B \(\sqrt{\frac{13}{19}}\)
- C \(\frac{\sqrt{19}}{3}\)
- D \(\frac{13}{\sqrt{19}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{19}{3}}\)
Step-by-step Solution
Detailed explanation
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 \(=2 a\)
\(\text { Length y con jugateaxis } =2 b=2 \sqrt{3} \)
\( e =\sqrt{\frac{1+b^{2}}{a^{2}}}=\sqrt{1+\left(\frac{4}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{1+16}{3}} \)
\( e =\sqrt{\frac{19}{3}}\)
\(
16\left(x^{2}-2 x\right)-3\left(y^{2}-4 y\right)=44
\)
\(
\begin{array}{l}
16(x+1)^{2}-3(y-2)^{2}=48 \\
\frac{(x+1)^{2}}{3}-\frac{(y-2)^{2}}{16}=1
\end{array}
\)
Centre \((1,-2)\) bength of ransuene axis \(=2 a\)
\(\text { Length y con jugateaxis } =2 b=2 \sqrt{3} \)
\( e =\sqrt{\frac{1+b^{2}}{a^{2}}}=\sqrt{1+\left(\frac{4}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{1+16}{3}} \)
\( e =\sqrt{\frac{19}{3}}\)
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