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:
$$\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^2+b^2}}$$

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! 
