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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?

  • $1/x$ is a hyperbola, and that is quite short – qwr Jul 26 '15 at 03:30
  • I think this is only half a parabola. For $a,b > 0$, you must have $x,y \ge 0$ or else you cannot take square roots. Maybe that's why that form is not more prevalent. – Marconius Jul 26 '15 at 03:58
  • It's not even half a parabola, just an arc. Here are two WolframAlpha links, with a plot and some parabolic properties. https://www.wolframalpha.com/input/?i=ContourPlot%5BSqrt%5Bx%5D%2BSqrt%5B2y%5D%3D%3D3%2C%7Bx%2C-1%2C10%7D%2C%7By%2C-1%2C10%7D%5D, https://www.wolframalpha.com/input/?i=Sqrt%5Bx%5D%2BSqrt%5B2y%5D%3D%3D3 –  Jul 26 '15 at 04:12
  • Using GeoGebra you can see the parabola and play with the parameters to view the changes. – Moti Jul 26 '15 at 05:16
  • To be more accurate - you get the portion of the parabola in the area of positive x and y. – Moti Jul 26 '15 at 05:22

2 Answers2

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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.

Will Jagy
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  • Thanks, Will. I was happy to recognize this as a parabola by the discriminant test, but it is interesting to see how the symbols expand with the transformations. –  Jul 26 '15 at 21:54
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Once the equation

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

in the earlier "rectangular parabola"

is recognized as representing a parabola with axis of symmetry inclined at $\pi/4$ to x-axis, touches the coordinate axes tangent at $x=a,y=a,$ it is not difficult to see that

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

represents a parabola with axis inclined at $\tan^{-1}(b/a)$ to x-axis touching the axes at $x=a,y=b.$

I had included a sketch of this particular situation in the above link.

Narasimham
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