<s>[INST] How to find the shortest distance from line to parabola? How to find the shortest distance from line to parabola?
parabola: $$2x^2-4xy+2y^2-x-y=0$$and the line is: $$9x-7y+16=0$$
Already tried use this formula for distance:
$$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$
 [/INST] $$2x^2-4xy+2y^2-x-y=0$$
At a point $(x_0, y_0)$ on the parabola, the gradient is 
$\nabla f(x_0,y_0) = (4x_0-4y_0-1, -4x_0+4y_0-1)$
The direction of a normal to the line $9x-7y+16=0$ is $(7,9)$. So we must have
\begin{align}
   (4x_0-4y_0-1, -4x_0+4y_0-1) \circ (7,9) &= 0 \\
   -8 x_0 + 8 y_0 - 16 &= 0 \\
   y_0 = 2+x_0 \\
\hline
   2 x_0^2 - 4 x_0 (2+x_0) + 2 (2+x_0)^2 - x_0 - (2+x_0)  &= 0 \\
   6 - 2x_0 &= 0 \\
   (x_0, y_0) &= (3, 5)
\end{align}
The distance from the line $9x-7y+16=0$ to the point $(3,5)$ is
$$\frac{|9x_0-7y_0+16|}{\sqrt{9^2+7^2}}
 =\frac{|27-35+16|}{\sqrt{130}} =\frac{8}{\sqrt{130}}$$
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