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Let $\mathrm{L}\left(\mathrm{x}_1, 4\right)$ be the end of the Latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ lying in the first quadrant and let $S\left(8, y_1\right)$ be the focus of the given hyperbola. Then the length of its transverse axis is
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Verified Answer
The correct answer is:
$4(\sqrt{17}-1)$
We know that end of the latus rectum in lying in the first quadrant be (ae, $\left.b^2 / a\right)$ and focus be $(c, 0)$ of hyperbola $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$
$\therefore$ According to question
$$
\frac{b^2}{a}=4 \Rightarrow b^2=4 a \ldots
$$
and $\mathrm{c}=8$
We have $c^2=a^2+b^2$
$$
\Rightarrow 64=a^2+49 \Rightarrow a^2+4 a-64=0
$$
$\begin{aligned} & \Rightarrow \mathrm{a}=\frac{-4 \pm \sqrt{272}}{2}=\frac{-4 \pm 4 \sqrt{17}}{2} \\ & \mathrm{a}=\frac{4 \sqrt{17}-4}{2} \text { [Take }+ \text { sign because end of latus rectium } \\ & \text { lies in first quadrant] } \\ & \therefore \text { Length of transverse axis }=2 \mathrm{a}=4(\sqrt{17}-1)\end{aligned}$
$\therefore$ According to question
$$
\frac{b^2}{a}=4 \Rightarrow b^2=4 a \ldots
$$
and $\mathrm{c}=8$
We have $c^2=a^2+b^2$
$$
\Rightarrow 64=a^2+49 \Rightarrow a^2+4 a-64=0
$$
$\begin{aligned} & \Rightarrow \mathrm{a}=\frac{-4 \pm \sqrt{272}}{2}=\frac{-4 \pm 4 \sqrt{17}}{2} \\ & \mathrm{a}=\frac{4 \sqrt{17}-4}{2} \text { [Take }+ \text { sign because end of latus rectium } \\ & \text { lies in first quadrant] } \\ & \therefore \text { Length of transverse axis }=2 \mathrm{a}=4(\sqrt{17}-1)\end{aligned}$
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