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what are the parts or the variables present in the bicorn equation?

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up vote 2 down vote accepted

According to the McGraw-Hill Dictionary of Scientific & Technical Terms, the bicorn curve is given by the solution set of $(x^2 + 2ay - a^2)^2 = y^2(a^2 - x^2)$, where $a$ is an arbitrary constant. The reference can be found here.

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because it is my research and , a is given in the problem and then it can now be graph! – user8562 Mar 22 '11 at 8:10
Can i ask another about the formula, a is the only variable right! if a is given then the equation can be derived! by the way can you help me, my professor wants me to have any info about the significance of area bisector of bicorne, well that my research topic,or bicorne itself but i cannot find any except the hat... can you help me please! – user8562 Mar 22 '11 at 8:12
I would say that $a$ was a scale constant and that $x$ and $y$ were the variables. The equation is a quadratic for $y$, so it is easy to draw, with two values of $y$ for each $x \in (-a,a)$. You can find more in Wikipedia – Henry Mar 22 '11 at 14:20

In geometry, the bicorn, also known as a cocked hat curve or bicorne is a rational quartic curve which has two cusps. The bicorn is a plane algebraic curve of degree four and genus zero. It has two cusp singularities in the real plane, and a double point in the complex projective plane at x=0, z=0.

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[Copied from the duplicate question]

Is the equation $$y^2(a^2-x^2)=(x^2+2ay-a^2)^2$$ (as suggested here) what you are trying to draw?

If so, it is a quadratic equation for $y$ which you can solve to something like

$$y = \frac{(a^2-x^2)\left( 2 a \pm \sqrt{ a^2 -x^2} \right)}{3 a^2 + x^2}$$

and then draw. You need $a^2 -x^2$ to be non-negative, and that gives you the range of $x$.

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I find it more convenient to use the parametric equations for the bicorn:

$$\begin{align*}x&=a\cos\,t\\y&=a\frac{\sin^2 t}{2+\sin\,t}\end{align*}$$

In Mathematica:

ParametricPlot[{Cos[t], Sin[t]^2/(2 + Sin[t])}, {t, 0, 2 Pi},
    Axes -> None, Frame -> True]


This checks that the equations are right:

y^2 (a^2 - x^2) == (x^2 + 2 a y - a^2)^2 /. 
  Thread[{x, y} -> {a Cos[t], a Sin[t]^2/(2 + Sin[t])}] // FullSimplify
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