<s>[INST] 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. 
 [/INST] 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,
$$\sqrt\frac{49}{32}<\sqrt{\frac{961}{320}}.$$
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