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. 

$$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$
so that is a couple of lines.
Hence you want to find the shortest distance from $y=-x$ or $y=2x$ to the parabola $y^2=x-2$.  Let this be denoted by the line segment with end points $(a, -a)$ or $(a, 2a)$  and $(b^2+2, b)$.
Then we need the min of $(a-b^2-2)^2+(a+b)^2$ or $(a-b^2-2)^2+(2a-b)^2$.
Case 1:
We have by Cauchy Schwarz
$$((a-b^2-2)^2+(a+b)^2)((-1)^2+1^2) \ge (-a+b^2+2+a+b)^2= ((b+\tfrac12)^2+\tfrac74)^2 \ge \frac{7^2}{4^2}$$
so we get a minimum distance of $\frac{7}{4\sqrt2}$.  
Case 2:
We have again by Cauchy Schwarz
$$((a-b^2-2)^2+(2a-b)^2)((-2)^2+1^2) \ge (-2a+2b^2+4+2a-b)^2= (2(b-\tfrac14)^2+\tfrac{31}8)^2 \ge \frac{31^2}{8^2}$$
so we get a minimum distance of $\frac{31}{8\sqrt5}$.  
As the second case happens to be higher, the shortest distance is $\dfrac7{4\sqrt2}$.
