Tangent line of Lissajous curve?

Question

I'm trying to find at how many points the tangent line of $(\cos(3t),\sin(2t))$ goes through the point $(3,0)$.

My attempt: This is the same thing as saying for how many values of $t$ do we have $x'(t)+3=0$ and $y'(t)=0$. So solving for $t$ in the system $(-k(3\sin(3t)+3), k 2\cos(2t))=(0,0)$. From the second equation we have $t=\frac{\pi}{4}+\pi m$. So we need to find when does $\sin(3t)=1$. So wouldn't $t=0,\frac{\pi}{6},\frac{5\pi}{6}$ in the interval $[0,2\pi]$.

But plotting the function shows there are at least four points where the tangent line intersects $(3,0)$. Did I make a mistake in my calculuation?

Answer 1

I plotted this, too, and confirm that there appear to be four tangents that pass through $(3,0)$.

Gradient of tangent is given by: $\frac {dy}{dx}=\frac {\frac {dy}{dt}}{\frac {dx}{dt}}=\frac {2 \cos 2t}{-3 \sin 3t}$

Equation of tangent is given by: $y=-\frac {2 \cos 2t}{3 \sin 3t}x+c$

Use the known point $\left(\cos 3t, \sin 2t \right)$ to find $c$

$\sin 2t=-\frac {2 \cos 2t}{3 \sin 3t}\cos 3t+c$

$c=\sin 2t+\frac {2 \cos 2t \cos 3t}{3 \sin 3t}$

Equation of tangent is therefore: $y=-\frac {2 \cos 2t}{3 \sin 3t}x+\sin 2t+\frac {2 \cos 2t \cos 3t}{3 \sin 3t}$

Set $x=3$ and $y=0$ to have:

$0=-\frac {2 \cos 2t}{\sin 3t}+\sin 2t+\frac {2 \cos 2t \cos 3t}{3 \sin 3t}$

$0=-6 \cos 2t+3 \sin 2t \sin 3t+2 \cos 2t \cos 3t$

Try using trig identities: $\sin 3t \equiv 3 \sin t - 4 \sin ^3 t$, $\cos 3t \equiv 4 \cos ^3 t -3 \cos t$, $\sin 2t \equiv 2 \sin t \cos t$, $\cos 2t \equiv 2 \cos ^2 t - 1$

$0=-12 \cos ^2 t + 6 +6 \sin t \cos t(3 \sin t - 4 \sin ^3 t)+2 (2 \cos ^2 t - 1)(4 \cos ^3 t -3 \cos t)$

$0=-12 \cos ^2 t + 6 + 18 (1-\cos^2 t)\cos t - 24 (1-\cos^2 t)^2 \cos t+16 \cos ^5 t - 20 \cos ^3 t +6 \cos t$

$0=-12 \cos ^2 t + 6 + 18 \cos t-18 \cos^3 t - 24 \cos t +48\cos^3 t-24\cos^5 t+16 \cos ^5 t - 20 \cos ^3 t +6 \cos t$

$0=-8 \cos ^5 t +10\cos^3 t -12 \cos ^2 t +6$

$0=4 \cos ^5 t -5 \cos^3 t + 6 \cos ^2 t - 3$

If $z=\cos t$, then $4z^5-5z^3+6z^2-3=0$

There are potentially five roots to this polynomial. A sketch shows that there are three roots.

Only two are suitable as a cosine, so that gives two values of $\cos t$. For each value of $\cos t$ there are two possible values of $t$, giving a total of four tangents.

Answer 2

Hint: For a given value of t, you can compute several quantities including x(t), y(t), x'(t), y'(t). Using these values, can you find the slope for the curve at a given value of t (hint: chain rule).

Since you then have the slope and a point on the tangent line (namely (x(t),y(t))), you can derive the formula for the tangent line.

Then, simply solve to see which values of t give you a line that passes through (3,0).