4

How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola. My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also for $\sqrt {x}-\sqrt {y}=1 $)

mathlover
  • 333
  • 1
    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Matthew Conroy Jan 28 '14 at 21:16
  • I rotated it in 45 degrees and I equation of parabola, but I don't know why this equation is only part of the parabola. – mathlover Jan 28 '14 at 21:17
  • You should add to your post the work you have done so others can see where to help you. – Matthew Conroy Jan 28 '14 at 21:18
  • It's only part of the parabola because the image of the square root function is the set of nonnegative reals; thus, $0 \le x, y \le 1$. – heropup Jan 28 '14 at 21:19
  • 2
    @recursiverecursion that is not correct. – heropup Jan 28 '14 at 21:22
  • You're right, sorry – recursive recursion Jan 28 '14 at 21:23

5 Answers5

4

From your equation, you get: $$y = (1-\sqrt{x})^2= 1-2\sqrt(x)+x$$ $$(x - y + 1)^2 = 4x$$ Now transform the variables: $$t = x+ y,\ \ s = x-y$$ $$(s+1)^2 = 2(s+t)$$ Or: $$t =\frac{s^2+1}{2}$$ Which is a parabola. This is only part of a parabola, since when $x$ or $y$ are smaller than zero, or larger than $1$, the original equation is undefined, since if either $x$ or $y$ were greater than one, the square root would have to return a negative value. This condition means your parabola is cropped by the square $[0,1]\times[0,1]$

Nathaniel Bubis
  • 33,018
3

Rearranging, we get $\sqrt{y} = 1 - \sqrt{x}$, which becomes $y = 1 - 2\sqrt{x} + x$ when we square both sides. Rearranging again and squaring both sides, we get $(y-x-1)^2 = y^2 + x^2 + 1 - 2xy - 2y + 2x = 4x$.

Generally, when there is an equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if $B^2 - 4AC = 0$, then the curve is one of the following:

  • A parabola
  • 2 Parallel Lines (certainly this is not the case)
  • 1 Line (certainly this is not the case)
  • No curve (certainly this is not the case)

Therefore, $\sqrt{x} + \sqrt{y} = 1$ is a part of a parabola.

(Credits partly go to WyzAnt)

2012ssohn
  • 3,827
3

Squaring, we get $x + y + 2\sqrt{xy} = 1$; moving the $2\sqrt{xy}$ term to one side, and squaring again, we obtain $$ x^2 -2xy + y^2 -x -y + 1 = 0$$

The value of the determinant $$\begin{vmatrix}A & B \\ B & C\\\end{vmatrix} $$ will tell us which type of curve is represented by a quadratic relation $$Ax^2 + 2Bxy + Cy^2 + Dx + Ey+F = 0;$$ we have a parabola if and only if the determinant is 0. (Details) Here $A=1, B=-1, C=1$, so the determinant is $\begin{vmatrix}1 & -1 \\ -1 & 1\\\end{vmatrix} = 0$, and this is indeed a parabola, truncated because it omits the points where $x<0$ or $y<0$.

MJD
  • 65,394
  • 39
  • 298
  • 580
  • nice answer @MJD – Semsem Jan 28 '14 at 21:29
  • 2
    It is truncated not because of the requirement that $x, y \ge 0$, but because $x, y \le 1$. To understand why, look at the derived conic section $P = {(x,y) \in \mathbb{R}^2 : x^2 - 2xy + y^2 - x - y + 1 = 0}$ that contains the subset $S = {(x,y) \in \mathbb{R}^2 : \sqrt{x} + \sqrt{y} = 1}$. $P$ itself does not contain any points such that $x < 0$ or $y < 0$. The truncation occurs when $x > 1$ or $y > 1$, as these are the points that are contained in $P$ but not in $S$. – heropup Jan 28 '14 at 21:45
1

Square both sides:

$$x + 2 \sqrt{xy} + y = 1.$$

Isolate the radical and square:

$$xy = \left(\frac{1-x-y}{2}\right)^2.$$

Rotate the axes by $\pi/4$ to new axes ${X,Y}$:

$$\frac{Y^2-X^2}{2} = \left(\frac{1 - \sqrt{2}Y}{2}\right)^2.$$

Simplfying gives:

$$X^2 = 2\sqrt{2}Y - 1,$$

which is the equation for a parabola.

John
  • 26,319
  • Why is this quation of parabola? – mathlover Jan 28 '14 at 21:42
  • It's a quadratic. – John Jan 28 '14 at 21:45
  • I like this answer more than the others because it's much more "short and sweet", so +1. But considering the level of the asker (and anyone else who finds this because they're searching for the answer to this question), the "rotate axes to get X and Y" part might need a bit more explanation. – Nike Dattani Nov 17 '20 at 21:49
0

Since the equation of this curve is $$x^2+y^2-2xy-2x-2y+1=0,$$ with $x,y\ge 0$, we can check that this is equivalent to $$ \sqrt{\left(x-\frac12\right)^2+\left(y-\frac12\right)^2}=\frac{x+y}{\sqrt2}. $$ In other words, the distance from $(x,y)$ to $\left(\frac12,\frac12\right)$ is equal to the distance from $(x,y)$ to the line $x+y=0$. Thus, $\sqrt{x}+\sqrt{y}=1$ is a parabola with focus $\left(\frac12,\frac12\right)$ and directrix $x+y=0$.

Alexander Burstein
  • 4,145