Is $\sqrt{x/a}+\sqrt{y/b}=1$ the equation of a parabola tangent to the coordinate axes? 
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.
 A: 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 ...


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.  
A: 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.)
A: 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.
A: 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 $y'=0$.
