Q. If the shortest distance between $2 y^2-2 x+1=0$ and $2 x^2-2 y+1=0$ is $d$, then find the number of solution of the equation $|\sin \theta|=2 \sqrt{2} d$ in the interval $[-\pi, 2 \pi]$.
Conic Sections
Solution:
Clearly, $d =\sqrt{\left(\frac{3}{4}-\frac{1}{2}\right)^2+\left(\frac{1}{2}-\frac{3}{4}\right)^2}=\sqrt{\frac{1}{16}+\frac{1}{16}}=\sqrt{\frac{1}{8}}=\frac{1}{2 \sqrt{2}}$
So, $|\sin \theta|=2 \sqrt{2} d \Rightarrow|\sin \theta|=1$
$\therefore \theta=\frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2} .$
Hence, the number of values of $\theta$ are 3 .
