Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ 
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $ab-h^2 =0 $  represent a parabola, 
$ab-h^2>1$ represent an ellipse 
$ab-h^2 <0$ represent a hyperbola 
Here $ab-h^2 <0$ therefore equation (1) represents hyperbola and equation (2) is parabola. ( where a,b, h are coefficients of $x^2,y^2,xy$ respectively.
Now how to get the minimum distance between the two curves please suggest .. thanks. 

When the shortest distance is achieved, the normal to the two curves coincide.
We are lucky that the first curve degenerates in $x+y=0$ and $2x-y=0$, giving two normal directions $(1,1)$ and $(2,-1)$.
Then, taking the gradient of the second expression, we express parallelism:
$$(-1,2y)\times(1,1)=-1-2y=0,$$
$$y=-\frac12,x=\frac94,$$
and compute the distance to the line
$$d=\frac{|\frac94-\frac12|}{\sqrt2}=\frac7{4\sqrt2}.$$ Similarly,
$$(-1,2y)\times(2,-1)=1-4y=0,$$
$$y=\frac14,x=\frac{33}{16},$$
$$d=\frac{|2\frac{33}{16}-\frac14|}{\sqrt5}=\frac{31}{8\sqrt5}.$$
Finally,
$$\sqrt\frac{49}{32}<\sqrt{\frac{961}{320}}.$$
