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. 

$y^2-xy-2x^2=0$, factorize it we get $(y-2x)(y+x)=0$, so the graph is two straight line $-2x+y=0$ or $x+y=0$.
Points on $y^2=x-2$ can be written as $(t^2+2,t)$, the distance from $(t^2+2,t)$ to $-2x+y=0$ is $\frac{|-2(t^2+2)+t|}{\sqrt{(-2)^2+1^2}}= \frac{|2t^2-t+4|}{\sqrt{5}}=\frac{|2(t-1/4)^2+31/8|}{\sqrt{5}}$.
The distance from $(t^2+2,t)$ to $x+y=0$ is $\frac{|t^2+2+t|}{\sqrt{(1)^2+1^2}}= \frac{|(t+1/2)^2+7/4|}{\sqrt{2}}$.
$\frac{7}{4\sqrt2}<\frac{31}{8\sqrt5}$ so $\frac{7}{4\sqrt2}=\frac{7\sqrt{2}}{8}$ is minimum distance.
