Is the below equation represents a parabola that touches the axes of coordinates? $$\sqrt{x/a}+\sqrt{y/b}=1$$

I know it is very stupid to ask this type of easy question here in the forum, but I'm very curious to know. I have searched many places and found nothing. My professor is not here, so I can't ask him. Suspense would have killed me.

Note from @Blue. Months later, I have edited the original problem to move the "$a$" and "$b$" under the radical signs. (This is because a duplicate problem recently appeared and I wanted to minimize confusion.) Most answers assumed this was the intention and proceeded accordingly. Those answers that use "$\sqrt{x}/a$" and "$\sqrt{y}/b$" should not be penalized for this after-the-fact notational change.

  • $\begingroup$ Do you know how to identify conics (in particular, parabolas) from the general second-degree form $A x^2 + B x y + C y^2 + D x + E y + F = 0$? Can you see how to transform your equation into second-degree form? $\endgroup$ – Blue Apr 9 '17 at 6:08
  • $\begingroup$ See revised solution using the Bezier Curve formula, which is much shorter and more direct. $\endgroup$ – hypergeometric Apr 17 '17 at 17:27

I'll take a different approach, describing a parabola that satisfies the equation. (More precisely, "a parabola with an arc that satisfies the equation", since, as @Alex notes, the equation's solution set must be bounded and therefore cannot include a complete parabola.)

Your original problem statement seemed a little unclear as to whether $a$ and $b$ belong inside the square roots. The first TeX edit of your question assumed they don't, and I preserved that assumption in my own edit. Here, however, I make the other call, so that the target is ... $$\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1 \tag{1}$$

where I'll take $a > 0$ and $b > 0$ (and therefore also $x > 0$ and $y > 0$). With that aside ...

enter image description here

My parabola is tangent to points $A=(a,0)$ and $B=(0,b)$. Its directrix, $\ell$, is perpendicular to diagonal $\overline{OC}$ of the rectangle $\square OACB$, and its focus, $F$, is the foot of the perpendicular from $O$ to $\overline{AB}$. Without too much trouble, we determine that the directrix has equation $$\ell : a x + b y = 0 \qquad\text{and}\qquad F = \frac{ab}{c^2}\left(b,a\right)$$ where $c := |\overline{OC}| = \sqrt{a^2+b^2}$. A point $(x,y)$ on the parabola must be equidistant to $F$ and $\ell$; invoking the corresponding distance formulas, we have ... $$\sqrt{\left(x-\frac{a b^2}{c^2}\right)^2 + \left(y-\frac{a^2 b}{c^2}\right)^2} = \frac{|a x + b y|}{c} \tag{2}$$ Squaring, clearing fractions, and expanding $c^2$ as $a^2 + b^2$, and then dividing-through by $a^2 b^2$, we can ultimately re-write the above as ... $$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 + \left(1\right)^2 - 2 \left(\frac{x}{a}\right)\left(\frac{y}{b}\right) - 2 \left(1\right) \left(\frac{x}{a}\right) - 2 \left(1\right) \left(\frac{y}{b}\right) = 0 \tag{3}$$ I've made various factors (and powers!) of $1$ conspicuous to put the reader in the mind of the expanded form of Heron's formula for the area of a triangle. Specifically, $(3)$ represents ($16$-times) the square of the area of a triangle with side-lengths $\sqrt{\frac{x}{a}}$, $\sqrt{\frac{y}{b}}$, $\sqrt{1}$. Since the area vanishes, we must have a degenerate "flat" triangle: two side-lengths must equal the third. The target equation $(1)$ represents one of the three ways this can happen, and its solution set is arc $\stackrel{\frown}{AB}$ of the parabola. The other cases, $$\sqrt{\frac{y}{b}} + 1 = \sqrt{\frac{x}{a}} \qquad\text{and}\qquad \sqrt{\frac{x}{a}} + 1 = \sqrt{\frac{y}{b}}$$ correspond to the unbounded "arms" attached at points $A$ and $B$, respectively. $\square$

It seems like it should be possible to make the degenerate triangle interpretation of $(3)$ "visible" in the diagram, but I have not yet found a good way to do this.

  • $\begingroup$ Should $F$ be $\displaystyle \frac {ab}{c^{\color{red}2}}\; (b,a)$? $\endgroup$ – hypergeometric Apr 17 '17 at 18:15
  • $\begingroup$ @hypergeometric: Indeed! Good eye ... Thanks. $\endgroup$ – Blue Apr 17 '17 at 22:41

(New Solution - much shorter and more direct!)

Here we adopt the form of the equation used in @Blue's solution, i.e. $$\boxed{\qquad \sqrt{\frac xa}+\sqrt{\frac yb}=1\qquad}$$ Converting to parametric form by putting $\displaystyle \sqrt{\frac xa}=t$ gives $$\begin{align} \left[x\atop y\right] &=\left[t^2 a\atop (1-t)^2 b\right]\\ &=(1-t)^2\left[0\atop b\right] +2(1-t)t\left[0\atop 0\right]+ t^2\left[a\atop 0\right] \end{align}$$ which is in the form of a Quadratic Bezier Curve (which is a parabola) with control points $B(0,b), \;O(0,0) ,\; A(a,0)$ where tangents to the curve at $B, A$ intersect at $O$.

Hence the equation represents (part of) a parabola which touches the $x-$axis and $y-$axis at points $A, B$ respectively. $\blacksquare$

(Previous Solution - much longer)

Taking the form used in @Blue's solution, we have

$$\begin{align} \sqrt{\frac xa}+\sqrt{\frac yb}&=1\tag{1}\\ \sqrt{bx}+\sqrt{ay}&=\sqrt{ab}\\ bx+ay+2\sqrt{abxy}&=ab\\ 4abxy&=\big[ab-(bx+ay)\big]^2\\ &=a^2b^2-2ab(bx+ay)+(bx+ay)^2\\ 0&=a^2b^2-2ab(bx+ay)+(bx-ay)^2\\ (bx-ay)^2&=2ab\left(bx+ay-\frac {ab}2\right)\tag{2}\\ (bx-ay)^2-2ab^2x-2a^2by+a^2b^2&=0\tag{3} \end{align}$$ As ($3$) is of the form $(Ax+Cy)^2+Dx+Ey+F=0$, it must be a parabola. See this.

Hence the curve ($1$) is part of the same parabola.

Note that ($2$) can also be written as

$$\left(\frac xa-\frac yb\right)^2=2\left(\frac xa+\frac yb\right)-1\tag{2a}$$ or $$\left(\frac xa-\frac yb-1\right)^2=\frac {4y}b\tag{2b}$$ It can also be worked out that the parabola touches the $x$ and $y$ axes at $A(a,0)$ and $B(0,b)$ respectively. Setting $x=0$ in ($2$) gives $(y-b)^2=0$ i.e. coincident roots at $y=b$. Similarly, setting $y=0$ in ($2$) gives $(x-a)^2=0$ i.e. coincident roots at $x=a$. Hence the coordinate axes are tangent to the parabola. $\blacksquare$

(Additional Notes)

Using information from the solutions here and here we can work out the following easily:

$$\begin{align} &\text{Parameter $t$:} &&t=\frac {ab(a^2-b^2)}{a^2+b^2}\\ &\text{Axis of symmetry:} &&bx-ay+\frac {ab(a^2-b^2)}{a^2+b^2}=0 &&\left[\frac xa-\frac yb+\frac {a^2-b^2}{a^2+b^2}=0\right]\\ &\text{Vertex, $V$, of parabola: } &&\left(\frac {ab^4}{(a^2+b^2)^2},\frac {a^4b}{(a^2+b^2)^2}\right)\\ &\text{Tangent at vertex:} &&ax+by-\frac{a^2b^2}{a^2+b^2}=0 &&\left[\frac xb+\frac ya-\frac {ab}{a^2+b^2}=0\right]\\ &\text{Directrix of parabola:} &&ax+by=0 &&\left[\frac xb+\frac ya=0\right]\\ &\text{Focus, $F$:} &&\left(\frac {ab^2}{a^2+b^2},\frac {a^2b}{a^2+b^2}\right)\\ &\text{Centre of Directrix*, $M$:} &&\left(\frac {ab^2(b^2-a^2)}{(a^2+b^2)^2},\frac {a^2b(a^2-b^2)}{(a^2+b^2)^2} \right)\\ &\text{Focal length, $z$:} &&\frac {a^2b^2}{(a^2+b^2)^{3/2}}=\frac {a^2b^2}{r^3} \end{align}$$

Note the following points:

  • *The centre of directrix, $M$, is the intersection between the axis of symmetry and the directrix. By definition, $FV=VM$.
  • The directrix is parallel to the tangent at the vertex.
  • $O$ lies on the directrix of the parabola. This is a standard property of the parabola - the intersection point of two perpendicular tangents to the parabola lies on its directrix.
  • The focus of the parabola, $F$, lies on the line $AB$ as well as the axis of symmetry.

See graphical implementation here.

enter image description here

(Further Addendum)

Note the following:

Using $r=\sqrt{a^2+b^2}$, and dividing the equations above by $r$,

the Axis of Symmetry (the "$Y$" axis) can also be written as $$\overbrace{\frac {bx-ay}{r}+\frac {ab(a^2-b^2)}{r^3}}^X=0$$ and the Tangent at the Vertex (the "$X$" axis) can also be written as

$$\overbrace{\frac {ax+by}r-\frac {a^2b^2}{r^3}}^Y=0$$

Using focal length $z=\dfrac{a^2b^2}r^3$, the equation of the parabola can then be written as

$$X^2=4zY\\ \color{red}{\left[\frac {bx-ay}{r}+\frac {ab(a^2-b^2)}{r^3}\right]^2=4\left(\frac{a^2b^2}{r^3}\right)\left[\frac {ax+by}r-\frac {a^2b^2}{r^3}\right]\tag{4}}$$ It can be shown that equation $(4)$ is equivalent to equations $(2), (2a), (2b), (3)$, and hence the complete parabola for $(1)$.

(Relationship with standard rotated parabola form)

Let vertex $\displaystyle V=(h,k)=\left(\frac{ab^4}{r^2},\frac{a^4b}{r^2}\right)$ where $r^2=a^2+b^2$ and $\displaystyle\tan\theta=\frac ba$.

Some pre-processing. Note that $$\color{orange}{\frac ha-\frac kb=\frac {b^4-a^4}{r^r}=\frac {(b^2-a^2)(b^2+a^2)}{(a^2+b^2)^2}=\frac {b^2-a^2}{r^2}}$$ and $$\color{green}{\frac hb+\frac ka=\frac {ab(a^2+b^2)}{(a^2+b^2)^2}=\frac {ab}{a^2+b^2}=\frac {ab}{r^2}}$$ Also, $$\color{blue}{-\frac {4a^2b^2}{r^4}-\left(\frac {b^2-a^2}{r^2}\right)^2=\frac {-4a^2b^2-(b^4-2a^2b^2+a^4)}{r^4}=\frac {-(b^2+a^2)^2}{r^4}=-1}$$. A parabola with focal length $\displaystyle z=\frac {a^2b^2}{r^3}$ with vertex at $V$ and axis of symmetry rotated by $\theta$ clockwise from the vertical is given by $$\begin{align} (x-h)\cos\theta+(h-k)\sin\theta &=\frac 1{4a}\big[(x-h)\sin\theta-(h-k)\cos\theta)\big]^2\\ (x-h)\frac ar+(y-k)\frac br &=\frac {r^3}{4a^2b^2}\big[(x-h)\frac br-(y-k)\frac ar\bigg]^2\\ \frac {ab}r\bigg[\frac {x-h}b+\frac {y-k}a\bigg] &=\frac {r^3}{4a^2b^2}\cdot \frac {a^2b^2}{r^2}\bigg[\frac {x-h}a-\frac {h-k}b\bigg]^2\\ \frac {4ab}{r^2}\bigg[\left(\frac xb+\frac ya\right)-\left(\color{orange}{\frac hb+\frac ka}\right)\bigg] &=\bigg[\left(\frac xa-\frac yb\right)-\left(\color{green}{\frac ha-\frac kb}\right)\bigg]^2\\ \frac {4ab}{r^2}\bigg[\left(\frac xb+\frac ya\right)-\color{orange}{\frac {ab}{r^2}}\bigg] &=\bigg[\left(\frac xa-\frac yb\right)-\color{green}{\frac {b^2-a^2}{r^2}}\bigg]^2\\ \frac {4ab}{r^2}\left(\frac xb+\frac ya\right)-\color{blue}{\frac {4a^2b^2}{r^4}} &=\left(\frac xa-\frac yb\right)^2-2\left(\frac {b^2-a^2}{r^2}\right)\left(\frac xa-\frac yb\right)+\color{blue}{\left(\frac {b^2-a^2}{r^2}\right)^2}\\ \left(\frac xa-\frac yb\right)^2 &=\frac {4ab}{r^2}\left(\frac xb+\frac ya\right)+2\left(\frac {b^2-a^2}{r^2}\right)\left(\frac xa-\frac yb\right)\color{blue}{-1}\\ &=\bigg[\frac {4a}{r^2}+\frac 2a\left(\frac{b^2-a^2}{r^2}\right)\bigg]x+\bigg[\frac {4b}{r^2}-\frac 2b\left(\frac{b^2-a^2}{r^2}\right)\bigg]y-1\\ &=\frac 2a\bigg[\frac {2a^2+b^2-a^2}{r^2}\bigg]x+\frac 2b\bigg[\frac {2b^2-(b^2-a^2)}{r^2}\bigg]y-1\\ &=\frac 2a\left(\frac {a^2+b^2}{\\ r^2}\right)x+\frac 2b\left(\frac{a^2+b^2}{r^2}\right)y-1\\ &=2\left(\frac xa+\frac yb\right)-1 \end{align}$$ which is effectively equation ($2a$) as derived from the original equation.

However, from the above, it can be seen working backwards from equation ($2a$) to the standard rotated/translated form is not quite so straightforward.

(From First Principles)

Start with the general equation for parabola, specify that it passes through and are tangential to the axes at $(a,0),(0,b)$.

General equation for parabola: $$(Ax+Cy)^2+Dx+Ey+F=0\tag{1}$$ At $(a,0):$ $A^2a^2+Da+F=0\tag{2}$ At $(0,b):$ $C^2b^2+Eb+F=0\tag{3}$ Differentiating $(1)$ and rearranging: $$\frac{dy}{dx}=-\frac {D+2A(Ax+CY)}{E+2C(aAx+Cy)}$$ At $(a,0)$, $\dfrac {dy}{dx}=0$ $\Rightarrow \quad D=-2A^2a\tag{4}$ At $(0,b)$, $\dfrac {dy}{dx}=\infty$ $\Rightarrow \quad E=-2C^2b\tag{5}$ Putting $(4),(5)$ in $(2),(3)$ gives $$F=A^2a^2=C^2b^2 \\ \Rightarrow {C=\pm \frac ab A\tag{6}}$$ Putting $(4),(5),(6)$ into $(1)$, diving by $A^2$ and rearranging: $$\left(\frac xa\pm\frac yb\right)^2-2\left(\frac xa+\frac yb\right)+1=0\tag{7}$$ Taking the $+$ sign in $\pm$ gives $$\left(\frac xa+\frac yb-1\right)^2=0$$ which graphs as two parallel lines.

Taking the $-$ sign in $\pm$ gives $$\left(\frac xa-\frac yb\right)^2=2\left(\frac xa+\frac yb\right)-1$$ which is the same as equation $(2a)$ derived from the original equation.

Hence the equation in the question represents part of a parabola to which the coordinate axes are tangential at $(a,0),(0,b)$ respectively.


Some further thoughts based on a refreshing method by a friend of mine who is an excellent mathematician.

First note that in parametric form the curve is $$\left(x\atop y\right)=\left(at^2\atop b(1-t)^2\right)$$ Apply the rotation matrix $\dfrac 1{\sqrt{a^2+b^2}}\left(\begin{array} \ b&-a\\a&\;\;b\end{array}\right)$ to get rid of $t^2$ in the $x$-component, i.e. rotating clockwise by $\arctan \left(\frac ab\right)$ about the origin: $$\begin{align} \left(X\atop Y\right) &=\frac 1{\sqrt{a^2+b^2}}\left(\begin{array} \ b&-a\\a&\;\;b\end{array}\right) \left(at^2\atop b(1-t)^2\right)\\ &=\frac 1{\sqrt{a^2+b^2}}\left(ab(2t-1)\atop (a^2+b^2)t^2-2b^2t+b^2\right)\qquad {\leftarrow \text{linear in $t$}\quad\;\atop {\leftarrow \text{quadratic in $t$}}}\\ &=\frac 1r\left(ab(2t-1)\atop r^2t^2-2b2t+b^2\right) \qquad\qquad\qquad\text{(where $r^2=a^2+b^2$)}\\ &=\frac 1r\left(ab(2t-1)\atop r^2\left(t-\frac {b^2}{r^2}\right)^2+\frac {a^2b^2}{a^2+b^2}\right) \end{align}$$ i.e. $Y=AX^2+BX+C$ which is a parabola. Hence the original curve is also a parabola. $\blacksquare$

By simple differentiation it can be shown that the the axes are tangent to the original parabola at $(a,0)$ and $(0,b)$. $\blacksquare$

Note that at $t=\frac {b^2}{r^2}$, $Y=Y_{\text{min}}=\frac {a^2b^2}{r^3}$ and $X=-\frac {ab(a^2-b^2)}{r^3}$, which is also the equation of the axis of symmetry.

Additional insights:

Using the fact that two perpendicular tangents (the coordinate axes in this case) to a parabola intersect at the directrix, we conclude that the origin $O$ lies on the directrix of the original parabola. Since $O$ is invariant under the applied (as the rotation is about $O$), therefore $O$ also lies on the directrix of the rotated parabola. Also, since the rotated parabola is upright, its directrix must be the $x-$axis itself. As such the focal length of the parabola must be $Y_{\text{min}}$, i.e. $\dfrac {a^2b^2}{r^3}=\dfrac {a^2b^2}{(a^2+b^2)^{3/2}}$.

Applying the reverse rotation matrix $\displaystyle\frac 1r\left(\;\;b\;\;a\atop -a\;\;b\right)$ to the vertex, axis of symmetry and directrix of the rotated parabola, it can be easily shown that, for the original parabola:

Vertex is $$\frac 1{(a^2+b^2)^2}\left(ab^4\atop a^4b\right)$$ Axis of symmetry is $$\left(\frac ar\left(\frac {b^2(a^2-b^2)}{r^3}+Y\right)\atop \frac br\left(-\frac {a^2(a^2-b^2)}{r^3}+Y\right)\right)\quad\Longrightarrow\quad \frac xa-\frac yb=\frac {a^2-b^2}{a^2+b^2}$$ Directrix is $$\frac 1r\left(\;\;bX\atop -aX\right)\quad\Longrightarrow\quad \frac xb+\frac ya=0$$

(Special Note)

See also this link here on the Superellipse.


You can see right away that your equation definitely does not represent an entire parabola.

The branches of a parabola go to infinity, whereas in your equation both $x$ and $y$ are bounded: $$ 0 \le x \le a^2, \qquad 0 \le y \le b^2. $$

However, if we transform the equation by squaring it, isolating the radical and squaring again (thus introducing into the picture infinitely many additional points $(x,y)$ that were not solutions of the original equation), then we would see that e.g. for $a=b=1$ your equation does represent a (bounded) subset of points of a parabola, $$4xy=(1-x-y)^2.\tag{1} $$

An easy way to see that the above equation describes a parabola is a linear transformation of coordinates: $x=v+u, \ y=v-u$; so $v={1\over2}(x+y), \ u={1\over2}(x-y)$. In the transformed coordinates, equation $(1)$ reduces to $$v = u^2 + {1\over4}, \tag{2} $$ so $v$ is a simple quadratic function of $u$.

In our $a=b=1$ example, the (arc of) parabola indeed touches the $x$ and $y$ axes at the points $(1,0)$ and $(0,1)$ because these two points correspond to the slope ${dv\over du}=\pm1$ at $u=\pm{1\over2}$ in the transformed coordinates $(u,v)$.

In the general case, for arbitrary positive $a$ and $b$, the touch points are $(a^2,0)$ and $(0,b^2)$. (We can go from the particular case $a=b=1$ to the general case simply by rescaling the coordinate axes. Such rescaling preserves the type of conic, so a parabola in rescaled axes remains a parabola.)

  • $\begingroup$ In the equation $$\frac{\sqrt{x}}{a}+\frac{\sqrt{y}}{b}=1,$$ which value of $\sqrt{x}$ should you take in order to get $\sqrt{y}=2b$? We must allow negative values of $\sqrt{\cdot}$ in order to get the entire parabola. My remark in the beginning should be understood "with positive (arithmetic) square root values". $\endgroup$ – Alex May 31 '17 at 2:20
  • $\begingroup$ In my answer I included a generalized ellipse. Only by squaring the entire parabola comes in. the given form only convex portion wrt arc comes in, as stated in the last line of my answer. $\endgroup$ – Narasimham May 31 '17 at 3:15

A lot of physical punch is lost from the parabola equation if we do not stick to dimensional agreement to see $a,b$ as segments made on the axes.

$$ \sqrt{x/a} + \sqrt{y/b} = 1\tag1$$

is a subset of generalized ellipse family

$$ \left(\frac{x}{a} \right)^n + \left(\frac{y}{b}\right)^n = 1 \tag2$$

When $\pm$ sign is respected we appreciate that it could belong to any of the four quadrants.

$$ \pm \sqrt{x/a} +\pm \sqrt{y/b} = 1\tag3$$

An equation of $\frac12$ order is not a conic. To get it to a classic conic form we need to massage it a bit.. to remove the $\pm$ in front of radicals we square two times, getting

$$ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2+1 =\pm 2 \frac{x}{a} \pm2 \frac{y}{b}\pm 2 \frac{xy}{ab} \tag4$$

The determinant $$ (2/ab)^2 - 4(1/a^2)(1/b)^2= 0$$ so they are all parabolas symmetrical to coordinate axes.

They are plotted below for values $ a= 3, b=2.$ It can be seen that the conics plot beyond tangent points which could not be enabled by the equation in radicals form.


To confirm tangency of the four parabolas to coordinate axes set $x=0$ or $y=0$ we see that $ x=a,y=b$ have double roots where the quadratic equations have zero discriminant and so are tangential to the $x,y$ coordinate axes.


Let it's possible for $x$-axes.

Hence, we have: $$\left(\frac{\sqrt{x}}{a}+\frac{\sqrt{y}}{b}\right)'=0$$ or $$\frac{1}{a\sqrt{x}}+\frac{y'}{b\sqrt{y}}=0,$$ which is impossible for $y'=0$.


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