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The locus of the point of intersection on the line $\sqrt{3} x-y-4 \sqrt{3} k=0$ and $\sqrt{3} k x+k y-4 \sqrt{3}=0$ for different real values of $k$ is a hyperbola $H$. If $e$ is the eccentricity of $H$, then $4 e^2=$
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Verified Answer
The correct answer is:
$16$
$$
\begin{aligned}
& \sqrt{3} x-y-4 \sqrt{3} k=0 \\
& \sqrt{3} k x+k y-4 \sqrt{3}=0
\end{aligned}
$$
Multiplying of $k$ in Eq. (i) and adding in Eq. (ii), we get
$$
x=2 \frac{\left(1+k^2\right)}{k}
$$
Multiplying of $k$ in Eq. (i) and subtracting from Eq. (ii), we get
$$
\begin{aligned}
& \quad \frac{y}{\sqrt{3}}=2\left(\frac{1-k^2}{k}\right) \\
& \therefore \quad x+\frac{y}{\sqrt{3}}=\frac{4}{k} \\
& \text { and } x-\frac{y}{\sqrt{3}}=4 k \\
& \Rightarrow \quad x^2-\frac{y^2}{3}=16
\end{aligned}
$$
$$
\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1
$$
Now, $e^2=1+\frac{48}{16}=4$
$$
\therefore \quad 4 e^2=16
$$
\begin{aligned}
& \sqrt{3} x-y-4 \sqrt{3} k=0 \\
& \sqrt{3} k x+k y-4 \sqrt{3}=0
\end{aligned}
$$
Multiplying of $k$ in Eq. (i) and adding in Eq. (ii), we get
$$
x=2 \frac{\left(1+k^2\right)}{k}
$$
Multiplying of $k$ in Eq. (i) and subtracting from Eq. (ii), we get
$$
\begin{aligned}
& \quad \frac{y}{\sqrt{3}}=2\left(\frac{1-k^2}{k}\right) \\
& \therefore \quad x+\frac{y}{\sqrt{3}}=\frac{4}{k} \\
& \text { and } x-\frac{y}{\sqrt{3}}=4 k \\
& \Rightarrow \quad x^2-\frac{y^2}{3}=16
\end{aligned}
$$
$$
\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1
$$
Now, $e^2=1+\frac{48}{16}=4$
$$
\therefore \quad 4 e^2=16
$$
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