Question

If the shortest distance between $2y^2-2x+1=0$ and $2x^2-2y+1=0$ is $d$, then find the number of solution of the equation $|\sin \theta |=2\sqrt{2}d$ in the interval $[-\pi ,2\pi ]$.

Answer 1

Answer: 3

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 .