Telling if an equation describes a parabola

Question

I am mostly convinced that the equation $y=x\pm2\sqrt x+1$ describes a parabola, but am not sure how to prove this. How would I do that?

Answer 1

Your equation is equivalent to $(y-x-1)^2=4x$.

Now, either you know how to get the conic type from the general equation $ax^2+bxy+cy^2+dx+ey+f=0$, and you develop the square and simplify. (see here)

Either you do an affine transformation $u=y-x-1, v=4x$, and in the new system of coordinates the equation is simply $v=u^2$, that of a parabola. Since an affine transformation does not change the type of conic, you are done.

Answer 2

First of all notice that your equation is only valid for $x \ge 0$.

We can re-arrange your equation to give \begin{eqnarray*} y &=& x\pm 2\sqrt{x}+1 \\ \\ y-x-1 &=& \pm 2 \sqrt x \\ \\ (y-x-1)^2 &=& 4x \\ \\ y^2+x^2+1-2xy-2y+2x&=& 4x \\ \\ x^2 + y^2 -2xy-2x-2y+1&=& 0 \end{eqnarray*}

The general form of a conic section is $ax^2 + by^2 + 2hxy + 2gx+2fy+c=0$ where $a,b,h,g,f,$ and $c$ are constants. Such a non-degenerate conic section is a parabola if, and only if, $h^2-ab=0$. (See below for an explanation)

In your case, $h=-1$, $a=1$ and $b=1$, meaning that $h^2-ab=(-1)^2-(1)(1)=0$.

It follows that both $y=x-2\sqrt x +1$ and $y=x+2\sqrt x + 1$ form subsets of parabolae.

Explanation: Conic sections are best seen as curves in projective space. In homogeneous coordinates $(x:y:z)$ the equation becomes $$ax^2+by^2+2hxy+2gxz+2fyz+cz^2=0$$ Ellipses miss the "line at infinite" $z=0$, parabolae are tangent to the line at infinity, and hyperbolae meet the line at infinity twice. Putting $z=0$ gives $ax^2+2hxy+by^2=0$. This has a repeated root if $h^2-ab=0$.