# There are many shapes of nail design. Can the “cut out” figure be drawn with mathematica graph?

I'd like to show you what the "cut out" design looks like.

Is it possible to plot a graph like this Plot[{8 x^2, 0.9 x^2 + 600}, {x, -10, 10}, AspectRatio -> 1] but with a single implicit formula? Or how do I plot a graph of two functions bounded only by interseptions?

• Try different powers of x, like e.g.: t = Table[Plot[(5 - y) x^2 + y x^6, {x, -1, 1}], {y, 0, 5, .5}]; Show[t] Commented Dec 23, 2020 at 8:39

(Not really a Mathematica question, but o well...)

An easy way is to use a piecewise combination of sine curves:

With[{a = 20, b = 5},
ParametricPlot[{10 Cos[t], ((b - a) UnitStep[t] - b) Sin[t]^2}, {t, -π, π}]]


where the two parameters control where the curve intersects the $$y$$-axis.

Just to explain where I got this from: I was inspired by the parametric equations of Sylvester's bicorn, and started modifying the equation for that accordingly. (Translate/rescale the resulting curve as seen fit.)

We can use BezierCurve to get the curve. And use Manipulate to get the control points.

a = {{-2, 0}, {-2, -8}, {2, -8}, {2, 0}};
Manipulate[
Show[Graphics[{Text[pts // TraditionalForm, {1, 1}], EdgeForm[White],
CMYKColor[4/100, 7/100, 19/100, 0, .5],
FilledCurve[{BSplineCurve[{{2, 0}, Sequence @@ pts, {-2, 0}},
SplineDegree -> 3], Line[a]}]}],
ParametricPlot[
BSplineFunction[{{2, 0}, Sequence @@ pts, {-2, 0}}][t], {t, 0, 1}],
PlotRange -> {{-2.5,
2.5}, {2, -10}}], {{pts, {{0.29, -1.44}, {-1.34, -2.32}}},
Locator}]


pts = {{0.29, -1.44}, {-1.34, -2.32}};
f = BSplineFunction[{{2, 0}, Sequence @@ pts, {-2, 0}}];
ParametricPlot[f[t], {t, 0, 1}]