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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Aicharya · Accepted Answer

$\begin{aligned} & 	ext {} \because 4 x^2-y^2-8 x-2 y-13=0 \ & \Rightarrow 4(x-1)^2-(y+1)^2=16 \ & \Rightarrow \frac{(x-1)^2}{4}-\frac{(y+1)^2}{16}=1 \end{aligned}$ Which can be parametrized as: $x=1+2 \sec 	heta, \quad y=-1+4 	an 	heta$ Now, $P\left(\frac{\pi}{4}ight)=\left(1+2 \sec \frac{\pi}{4},-1+4 	an \frac{\pi}{4}ight)$ $\begin{gathered} P\left(\frac{\pi}{4}ight)=(1+2 \sqrt{2}, 3) \ Q\left(\frac{3 \pi}{4}ight)=\left(1+2 \sec \left(\frac{3 \pi}{4}ight),-1+4 	an \left(\frac{3 \pi}{4}ight)ight)=(1-2 \sqrt{2},-5) \end{gathered}$ Distance between $P Q=\sqrt{(1+2 \sqrt{2}-1+2 \sqrt{2})^2+(-3-5)^2}$ $=\sqrt{32+64}=\sqrt{96}=4 \sqrt{6}$

If $\mathrm{P}\left(\frac{\pi}{4}\right)$ and $\mathrm{Q}\left(\frac{3 \pi}{4}\right)$ are two points on the…

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