(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.
(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.
(ANOTHER METHOD)
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.
