How to Graph Parametric Equations Using a TI Calculator
How to Graph Parametric Equations Using a TI Calculator

## Third solution

A slightly simpler and more geometric approach leads to a third form for the solution (including Artes’ and my second) — don’t you just love trigonometric functions!

By symmetry, two points starting from a vertex of the hypocycloid (star) and going in opposite directions at the same speed will meet at one of the desired crossings.

If the points start at {4, 0}, then at time t the parameters will be t, 4 Pi - t.
The crossing times of the points can be found by setting the parametrizations equal to each other and solving.

center = {(12 Cos[t])/5, (12 Sin[t])/5}; radius = {8/5 Cos[(3 t)/2], -(8/5) Sin[(3 t)/2]}; param = center + radius; sol0 = {t, 4 Pi - t} /. First@Solve[param == (param /. t -> 4 Pi - t) && 0 < t < 2 Pi, t] (* {2 ArcCos[-(1/4)], 4 Pi - 2 ArcCos[-(1/4)]} *) 

To get the others, add multiples of 1/5 of the period 4 Pi:

Table[Mod[Join[#, N@#] &@ (sol0 + 4 Pi/5 i), 4 Pi], {i, 0, 4}] // Grid 

$$\begin{array}{cc|cc} t_1 & t_2 & N[t_1] & N[t_2]\strut \\ \hline 2 \cos ^{-1}\left(-{1}/{4}\right) & 4 \pi -2 \cos ^{-1}\left(-{1}/{4}\right) & 3.64695 & 8.91942 \\ \textstyle \frac{4 \pi }{5}+2 \cos ^{-1}\left(-{1}/{4}\right) & \textstyle \frac{24 \pi }{5}-2 \cos ^{-1}\left(-{1}/{4}\right) & 6.16023 & 11.4327 \\ \textstyle \frac{8 \pi }{5}+2 \cos ^{-1}\left(-{1}/{4}\right) & \textstyle \frac{8 \pi }{5}-2 \cos ^{-1}\left(-{1}/{4}\right) & 8.6735 & 1.3796 \\ \textstyle \frac{12 \pi }{5}+2 \cos ^{-1}\left(-{1}/{4}\right) & \textstyle \frac{12 \pi }{5}-2 \cos ^{-1}\left(-{1}/{4}\right) & 11.1868 & 3.89287 \\ \textstyle-\frac{4 \pi }{5} + 2 \cos ^{-1}\left(-{1}/{4}\right) & \textstyle \frac{16 \pi }{5}-2 \cos ^{-1}\left(-{1}/{4}\right) & 1.13368 & 6.40614 \\ \end{array}$$

The equations are not hard to solve by hand either. Setting $t = 2\theta$, the condition $y = 0$ reduces to $2 \sin 3\theta = 3 \sin 2\theta$. Applying multiple-angle identities, one discovers the factor $1 + 4 \cos\theta$, from which the solution follows.

Here a movie illustrating the method:

hypoc = ParametricPlot[param, {t, 0, 4 Pi}, PlotRange -> 4.05]; wheel = {Circle[center, 8/5], PointSize[Large], Point[param], Line[{{center, param}, {{0, 0}, center}}], Opacity[0.5], Line@Table[center + # & /@ ({radius, 0.9 radius} /. t -> t - 2 dt/3), {dt, Pi/8, 2 Pi - Pi/8, Pi/8}]}; Manipulate[ Show[hypoc, Graphics@Dynamic[{ Circle[{0, 0}, 4], Red, wheel /. t -> t + offset, Darker@Green, wheel /. t -> 4 Pi - t + offset } /. t -> t0] ], {{t0, First@sol0}, 0, 4 Pi, First@sol0/100}, {offset, 4 Pi/5 Range[0, 4]} ] 

## Second solution

Knowing it’s class work made me think along simpler lines than at the beginning.

param = {12/5 Cos[t] + 8/5 Cos[(3 t)/2], 12/5 Sin[t] - 8/5 Sin[(3 t)/2]}; sol = Solve[param[[2]] == 0, {t}] (* {{t -> ConditionalExpression[4 \[Pi] C[1], C[1] \[Element] Integers]}, {t -> ConditionalExpression[2 (\[Pi] + 2 \[Pi] C[1]), C[1] \[Element] Integers]}, {t -> ConditionalExpression[2 (\[Pi] - ArcTan[Sqrt[15]] + 2 \[Pi] C[1]), C[1] \[Element] Integers]}, {t -> ConditionalExpression[2 (-\[Pi] + ArcTan[Sqrt[15]] + 2 \[Pi] C[1]), C[1] \[Element] Integers]}} *) 

The third solution is one of the crossings:

param /. sol /. {C[1] -> 0} // N {{4., 0.}, {0.8, 0.}, {-1., 1.33227*10^-15}, {-1., -1.33227*10^-15}} 

The next crossing will be 3 * 4 Pi / 5 further along.

{t, t + 3 * 4 Pi/5} /. sol[[3]] /. {C[1] -> 0} (* {2 (Pi - ArcTan[Sqrt[15]]), (12 Pi)/5 + 2 (Pi - ArcTan[Sqrt[15]])} *) % // N (* {3.64695, 11.1868} *) 

The points are passed through again, by symmetry, at these values of t:

{4 Pi, 4 Pi + 4 Pi/5} - % (* {8.91942, 3.89287} *) 

All intersections:

t1 = t + Range[0, 4 Pi - 1, 4 Pi/5] /. sol[[3]] /. {C[1] -> 0}; t2 = {4 Pi, 4 Pi, 0, 0, 0} + Range[0, 2 4 Pi - 1, 2 4 Pi/5] - t1; Transpose[{t1, t2}] // Expand // Grid 

$\begin{array}{cc} 2 \pi -2 \tan ^{-1}\left(\sqrt{15}\right) & 2 \pi +2 \tan ^{-1}\left(\sqrt{15}\right) \\ \frac{14 \pi }{5}-2 \tan ^{-1}\left(\sqrt{15}\right) & \frac{14 \pi }{5}+2 \tan ^{-1}\left(\sqrt{15}\right) \\ \frac{18 \pi }{5}-2 \tan ^{-1}\left(\sqrt{15}\right) & \frac{18 \pi }{5}+2 \tan ^{-1}\left(\sqrt{15}\right) \\ \frac{22 \pi }{5}-2 \tan ^{-1}\left(\sqrt{15}\right) & \frac{22 \pi }{5}+2 \tan ^{-1}\left(\sqrt{15}\right) \\ \frac{26 \pi }{5}-2 \tan ^{-1}\left(\sqrt{15}\right) & \frac{26 \pi }{5}+2 \tan ^{-1}\left(\sqrt{15}\right) \\ \end{array}$

Original solution

[This method is a good way to find roots when the equations cannot be solve explicitly, or Solve/Reduce take a very long time.]

You can record the t values of the plot with EvaluationMonitor, Reap, and Sow.

{plot, {pts}} = Reap @ ParametricPlot[{2.4*Cos[t] + 1.6*Cos[3 t/2], 2.4*Sin[t] - 1.6 Sin[3 t/2]}, {t, 0, 4*Pi}, EvaluationMonitor :> Sow[{t, 2.4*Cos[t] + 1.6*Cos[3 t/2], 2.4*Sin[t] - 1.6 Sin[3 t/2]}]]; 

Then we can construct a NearestFunction to find the values of t that produce points closest to the approximate crossing.

nf = Nearest[pts[[All, 2 ;; 3]] -> pts[[All, 1]]] (* NearestFunction[{526, 2}, <>] *) crossings = {{-0.3054, -0.9543}, {-1.007, -0.005534}}; (* {{11.1513, 3.8396}, {8.96306, 3.57175}} *) 

Each approximate crossing produces two distinct values for t

nf[#, 2] & /@ crossings (* {{11.1513, 3.8396}, {8.96306, 3.57175}} *) 

We can set up starting points for FindRoot like this:

Transpose@{{t, s}, nf[crossings[[1]], 2]} (* {{t, 11.1513}, {s, 3.8396}} *) 

Then FindRoot will refine the estimates:

param = {2.4*Cos[t] + 1.6*Cos[3 t/2], 2.4*Sin[t] - 1.6 Sin[3 t/2]}; sol1 = FindRoot[param - (param /. t -> s), Transpose@{{t, s}, nf[crossings[[1]], 2]}] (* {t -> 11.1868, s -> 3.89287} *) sol2 = FindRoot[param - (param /. t -> s), Transpose@{{t, s}, nf[crossings[[2]], 2]}] (* {t -> 8.91942, s -> 3.64695} *) Show[plot, Graphics[{PointSize[Large], Red, Point[param /. {sol1, sol2}]}]] 

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