Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$

Question

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.

Answer 1

$$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$ 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 $\dfrac7{4\sqrt2}$.

Answer 2

Let $$ f(x,y) = y^2-xy-2x^2\\ g(x,y) = y^2-x+2. $$ The segment of minimum distance has the direction of the normal line to both the conics. The normal line has the direction of the gradient. Hence you could try to solve the following system: $$ \begin{cases} \nabla f((x,y) + \lambda \nabla g(x,y)) = \mu\nabla g(x,y)\\ f((x,y)+\lambda\nabla g(x,y)) = 0\\ g(x,y) = 0 \end{cases} $$ where $$ \nabla f(x,y) = (-y-4x, 2y-x)\\ \nabla g(x,y) = (-1,2y) $$

Answer 3

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, $$\sqrt\frac{49}{32}<\sqrt{\frac{961}{320}}.$$

Answer 4

$y^2-xy-2x^2=0$, 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.

Answer 5

The first conic is two straight lines, of which the one which is closest to the parabola is the line $x+y=0$. So you just need to find the point P on the parabola where the gradient is $-1$, and find the point where the normal at P meets this line.