Does anyone have a reference on the equation $$\sqrt{ax}\,+\sqrt{by}=c\ ?$$
Clearing square roots and rearranging gives $$ax+by = \frac{(ax-by)^2+c^4}{2c^2}$$
This is the equation of a parabola, so the original equation is a parabolic arc.
I'm surprised because I've never seen this before: I've never seen such a short equation yield a parabola at an angle from the coordinate axes. Does anyone have any reference discussing this?
no reference... I cannot tell whether you know how to do this. EDIT: judging from your MO answers, you do. Would have been better to include this in your question....
Rotated coordinate system by $$ u = \frac{ax-by}{\sqrt {a^2 + b^2}}, $$ $$ v = \frac{bx+ay}{\sqrt {a^2 + b^2}}, $$ $$ x = \frac{au+bv}{\sqrt {a^2 + b^2}}, $$ $$ y = \frac{-bu+av}{\sqrt {a^2 + b^2}}. $$
In particular $$ ax+by = \frac{(a^2 - b^2)u + 2abv}{\sqrt {a^2 + b^2}}. $$ Let us add in $$ ax-by = \left(\sqrt {a^2 + b^2}\right) u. $$
Taking your final formula, I get $$ (a^2-b^2)u + 2ab v = \frac{\sqrt {a^2 + b^2}}{2c^2} \left( (a^2 + b^2)u^2 + c^4 \right) $$ and one may solve for $v$ as some $$ v = E u^2 + F u + G $$ which is a parabola, as advertised. The original solution set is the subset of this that is close to the origina and stops at the two points where the parabola is tangent to the $x$ and $y$ axes.
Lat night, I forgot one factor of $2,$ which is why I got the wrong conclusion. It became much clearer after I did the symmetric example $\sqrt x + \sqrt y = 1,$ which is rotated exactly $45^\circ.$ So, I suggest including one or two examples, typeset, so as to offset the influence of too many symbols.