parabola

\hspace{-16}$Here Two Parabola $\mathbf{2y^2=2x-1}$ and $\mathbf{2x^2=2y-1}$\\\\ are Inverse of each other along The line $\mathbf{y=x}$\\\\ Now Shortest Distance b/w These Two parabola are possible only When\\\\ Slope of Tangents at any points on These Parabola are same OR Same as\\\\ slope of line $\mathbf{y=x}$\\\\ So Slope of Tangents at points $\mathbf{(x_{1},y_{1})}$ on $\mathbf{2x^2=2y-1}$ is $\mathbf{\frac{dy}{dx}\Bigint|_(x_{1},y_{1})=2x_{1}}$ and slope of $y=x$ is $\mathbf{\frac{dy}{dx}=1}$\\\\ So $\mathbf{2x_{1}=1\Leftrightarrow x_{1}=\frac{1}{2}}$ and $\mathbf{y_{1}=\frac{3}{4}}$.\\\\ Now These point lie on The parabola $\mathbf{2x^2=2y-1}$.\\\\ Now Draw Perpendicular $\mathbf{A\left(\frac{1}{2},\frac{3}{4}\right)}$ to line $\mathbf{x-y=0}$\\\\ So Length of Perpendicular $\mathbf{AC=\left|\frac{\frac{1}{2}-\frac{3}{4}}{\sqrt{2}}\right|=\frac{1}{4\sqrt{2}}}$\\\\ So $\mathbf{AB=2AC=\frac{1}{2.\sqrt{2}}}$..