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If $\mathrm{P}\left(\frac{\pi}{4}\right)$ and $\mathrm{Q}\left(\frac{3 \pi}{4}\right)$ are two points on the hyperbola $4 x^2-y^2-8 x-2 y-13=0$ in parametric form, then the distance between $\mathrm{P}$ and $\mathrm{Q}$ is
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Verified Answer
The correct answer is:
$4 \sqrt{6}$
$\begin{aligned}
& \text {} \because 4 x^2-y^2-8 x-2 y-13=0 \\
& \Rightarrow 4(x-1)^2-(y+1)^2=16 \\
& \Rightarrow \frac{(x-1)^2}{4}-\frac{(y+1)^2}{16}=1
\end{aligned}$
Which can be parametrized as:
$x=1+2 \sec \theta, \quad y=-1+4 \tan \theta$
Now, $P\left(\frac{\pi}{4}\right)=\left(1+2 \sec \frac{\pi}{4},-1+4 \tan \frac{\pi}{4}\right)$
$\begin{gathered}
P\left(\frac{\pi}{4}\right)=(1+2 \sqrt{2}, 3) \\
Q\left(\frac{3 \pi}{4}\right)=\left(1+2 \sec \left(\frac{3 \pi}{4}\right),-1+4 \tan \left(\frac{3 \pi}{4}\right)\right)=(1-2 \sqrt{2},-5)
\end{gathered}$
Distance between $P Q=\sqrt{(1+2 \sqrt{2}-1+2 \sqrt{2})^2+(-3-5)^2}$ $=\sqrt{32+64}=\sqrt{96}=4 \sqrt{6}$
& \text {} \because 4 x^2-y^2-8 x-2 y-13=0 \\
& \Rightarrow 4(x-1)^2-(y+1)^2=16 \\
& \Rightarrow \frac{(x-1)^2}{4}-\frac{(y+1)^2}{16}=1
\end{aligned}$
Which can be parametrized as:
$x=1+2 \sec \theta, \quad y=-1+4 \tan \theta$
Now, $P\left(\frac{\pi}{4}\right)=\left(1+2 \sec \frac{\pi}{4},-1+4 \tan \frac{\pi}{4}\right)$
$\begin{gathered}
P\left(\frac{\pi}{4}\right)=(1+2 \sqrt{2}, 3) \\
Q\left(\frac{3 \pi}{4}\right)=\left(1+2 \sec \left(\frac{3 \pi}{4}\right),-1+4 \tan \left(\frac{3 \pi}{4}\right)\right)=(1-2 \sqrt{2},-5)
\end{gathered}$
Distance between $P Q=\sqrt{(1+2 \sqrt{2}-1+2 \sqrt{2})^2+(-3-5)^2}$ $=\sqrt{32+64}=\sqrt{96}=4 \sqrt{6}$
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