Let $p(x_0, y_0)$ be the point hyperbola $3x^2-4y^2=36$ which is nearest to the line $3x+2y=1$ Then $\sqrt{2}(y_0-x_0)$ is equal to ______

$3x^2-4y^2=363x+2y=1$

$m=-\frac{3}{2}$

$m=+\frac{3\sec \theta }{\sqrt{12}.\tan \theta }\Rightarrow \frac{3}{\sqrt{12}}\times \frac{1}{\sin \theta }=\frac{-3}{2}$

$\sin \theta =-\frac{1}{\sqrt{3}}$

$(\sqrt{12}\sec \theta ,3\tan \theta )$

$(\sqrt{12}\frac{\sqrt{3}}{\sqrt{2}},-3\times \frac{1}{\sqrt{2}})\Rightarrow (\frac{6}{\sqrt{2}},\frac{-3}{\sqrt{2}})$