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If the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,(a>b)$ has eccentricity and the length of the latusrectum respectively equal to $\frac{5}{4}$ and 9 , then $a b$ is equal to
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Verified Answer
The correct answer is:
48
Eccentricity $(e)=\sqrt{1+\frac{b^2}{a^2}}$
$$
\begin{aligned}
\Rightarrow & 1+\frac{b^2}{a^2} =\frac{25}{16} \quad given \\
\Rightarrow & b^2 =\frac{9}{16} a^2
\end{aligned}
$$
and length of latus rectum $=9$(given)
$$
\begin{aligned}
\Rightarrow & & \frac{2 b^2}{a} & =9 \Rightarrow b^2=\frac{9 a}{2} \\
& \Rightarrow & \frac{9}{16} a^2 & =\frac{9 a}{2} \Rightarrow a=8 \\
& \therefore & b^2 & =\frac{9}{16} \times(8)^2=36 \\
& \Rightarrow & b & =6 \\
& \therefore & a b & =48
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow & 1+\frac{b^2}{a^2} =\frac{25}{16} \quad given \\
\Rightarrow & b^2 =\frac{9}{16} a^2
\end{aligned}
$$
and length of latus rectum $=9$(given)
$$
\begin{aligned}
\Rightarrow & & \frac{2 b^2}{a} & =9 \Rightarrow b^2=\frac{9 a}{2} \\
& \Rightarrow & \frac{9}{16} a^2 & =\frac{9 a}{2} \Rightarrow a=8 \\
& \therefore & b^2 & =\frac{9}{16} \times(8)^2=36 \\
& \Rightarrow & b & =6 \\
& \therefore & a b & =48
\end{aligned}
$$
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