# How to Draw the Famous Batman Curve

The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:

So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$:

That’s the first factor.

The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:

This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:

and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:

The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines $y = 9 - 8|x|$:

truncated to the region $0.75 < |x| < 1$.

Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| \times \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$:

truncated to the region $0.5 < |x| < 0.75$.

The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.

Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:

so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like

As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving:

The MATLAB M-File is :

clf; clc; clear all; syms x y eq1 = ((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1); eq2 = (abs(x/2)-((3*sqrt(33)-7)/112)*x^2-3+sqrt(1-(abs(abs(x)-2)-1)^2)-y); eq3 = (9*sqrt(abs((abs(x)-1)*(abs(x)-.75))/((1-abs(x))*(abs(x)-.75)))-8*abs(x)-y); eq4 = (3*abs(x)+.75*sqrt(abs((abs(x)-.75)*(abs(x)-.5))/((.75-abs(x))*(abs(x)-.5)))-y); eq5 = (2.25*sqrt(abs((x-.5)*(x+.5))/((.5-x)*(.5+x)))-y); eq6 = (6*sqrt(10)/7+(1.5-.5*abs(x))*sqrt(abs(abs(x)-1)/(abs(x)-1))-(6*sqrt(10)/14)*sqrt(4-(abs(x)-1)^2)-y); axes('Xlim', [-7.25 7.25], 'Ylim', [-5 5]); hold on ezplot(eq1,[-8 8 -3*sqrt(33)/7 6-4*sqrt(33)/7]); ezplot(eq2,[-4 4]); ezplot(eq3,[-1 -0.75 -5 5]); ezplot(eq3,[0.75 1 -5 5]); ezplot(eq4,[-0.75 0.75 2.25 5]); ezplot(eq5,[-0.5 0.5 -5 5]); ezplot(eq6,[-3 -1 -5 5]); ezplot(eq6,[1 3 -5 5]); colormap([0 0 1]) title('Batman'); xlabel(''); ylabel(''); hold off

Source:
The Batman Equation
(math.stackexchange.com)