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:
$$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$
 A: hint
The parametric equations are
for parabola
$$2 (x-y)^2=x+y $$
$$x-y=t $$
$$x+y=2t^2$$
thus

$$x=t^2+t/2 \;,\;y=t^2-t/2$$

the   distance from a point of parabola to  the line is 
$$D=\frac {|9 (t^2+t/2)-7 (t^2-t/2)+16|}{\sqrt{81+49}} $$
$$=\frac {2t^2+8t+16}{\sqrt {130}} $$
the minimum is attained for $t=-2$ and it is

$$D_{min}=\frac{8}{\sqrt {130} }$$

A: Look at the graphs:

The tangent line to the parabola at the point $(x_0,y_0)$ is: $y=y_0+y'(x_0)(x-x_0).$
The tangent line must be parallel to the line, hence: $y'(x_0)=\frac97 \ \  \ (1)$
Take implicit differentiation from the equation of the parabola and solve the equation $(1)$ to find the point $(x_0,y_0).$
Then you find the distance between the point $(x_0,y_0)$ and the line.
A: HINT:
Use rotation of axes to eliminate the $xy$ term from the equation of the parabola as the distance is invariant in rotation.
Now use parametric equation $P(h+at^2,k+2at)$ of the parabola $$(y-k)^2=4a(x-h)$$ and $$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$
A: We need to find the minimum of $$\frac{|9x-7y+16|}{\sqrt{9^2+7^2}},$$
where $x+y=2(x-y)^2$ or
$$\min\frac{|(9x-7y)(x+y)+32(x-y)^2|}{2\sqrt{130}(x-y)^2}$$ or
$$\min\frac{|41x^2-62xy+25y^2|}{2\sqrt{130}(x-y)^2}$$ or
$$\min\frac{41x^2-62xy+25y^2}{2\sqrt{130}(x-y)^2},$$
which is $\frac{8}{\sqrt{130}}$ because
$$\frac{41x^2-62xy+25y^2}{2\sqrt{130}(x-y)^2}\geq\frac{8}{\sqrt{130}}$$
it's just $$(5x-3y)^2\geq0$$ and the equality occurs.
Done! 
A: $$2x^2-4xy+2y^2-x-y=0$$
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
$$\frac{|9x_0-7y_0+16|}{\sqrt{9^2+7^2}}
 =\frac{|27-35+16|}{\sqrt{130}} =\frac{8}{\sqrt{130}}$$
