Question

The curve x = 2 cos(t), y = sin(t) − sin(2t), 0 ≤ t ≤ 2π crosses itself once,find this intersection point and then find the equations of the two tangent lines at the intersection point.

Answer 1

You're looking for
`t_1\\eq t_2` such that
`(x(t_1),y(t_1))=(x(t_2),y(t_2))`.


`\begin{cases}2\cos t_1=2\cos t_2\\\sin t_1-\sin2t_1=\sin t_2-\sin2t_2\end{cases}`

Recall that
`\sin2x=2\sin x\cos x`, so the second equation can be written as


`\sin t_1-2\sin t_1\cos t_1=\sin t_2-2\sin t_2\cos t_2`

`\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_2)`

Since
`2\cos t_1=2\cos 2_t`, and assuming
`1-2\cos t_2\\eq0`, you get


`\sin t_1(1-2\cos t_1)=\sint t_2(1-2\cos t_1)`

`(\sin t_1-\sin t_2)(1-2\cos t_1)=0`

which admits two possibilities; either
`\sin t_1=\sin t_2` or
`1-2\cos t_1=0`. In the first case, since we're assuming
`t_1\\eq t_2`, we can use the fact that
`\sin(\pi-x)=\sin x` to arrive at a solution of
`t_2=\pi-t_1`.

In the second case, you have


`1-2\cos t_1=0\implies \cos t_1=\frac12\implies t_1=\frac\pi3\text{ or }\frac{5\pi}3`

Let's check which of these solutions work. If
`t_1=\frac\pi3`, then the sine equation suggests
`t_2=\pi-\frac\pi3=\frac{2\pi}3`. However,


`\left(x\left(\frac\pi3\right),y\left(\frac\pi3\right)\right)=(1,0)`

`\left(x\left(\frac{2\pi}3\right),y\left(\frac{2\pi}3\right)\right)=(-1,\sqrt3)`

so in fact this is an extraneous solution. So let's return to the first equation in the system,


`2\cos t_1=2\cos t_2\implies \cos t_1=\cos t_2`

Again, assuming
`t_1\\eq t_2`, we can use the fact that
`\cos(2\pi-x)=\cos x` to arrive at a solution of
`t_2=2\pi-t_2`. Now, if
`t_1=\frac\pi3`, we get
`t_2=\frac{5\pi}3`. Let's check if this works:


`\left(x\left(\frac\pi3\right),y\left(\frac\pi3\right)\right)=(1,0)`

`\left(x\left(\frac{5\pi}3\right),y\left(\frac{5\pi}3\right)\right)=(1,0)`

Indeed, this solution works! So the curve intersects itself at the point (1,0), which the curve passes for the first time through when
`t=\frac\pi3` and the second time when
`t=\frac{5\pi}3`.

Now, to find the tangent line, we need to compute the derivative of
`y` with respect to
`x`. You have


`(\mathrm dy)/(\mathrm dx)=((\mathrm dy)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))=(\cos t-2\cos2t)/(-2\sin t)=(2\cos2t-\cos t)/(2\sin t)`

When
`t=\frac\pi3`, you have a slope of
`-\frac{\sqrt3}2`; at
`t=\frac{5\pi}3`, the slope is
`\frac{\sqrt3}2`.

The tangent lines are then


`y_1-0=-\frac{\sqrt3}2(x-1)\implies y_1=-\frac{\sqrt3}2x+\frac{\sqrt3}2`

`y_2-0=\frac{\sqrt3}2(x-1)\implies y_2=\frac{\sqrt3}2x-\frac{\sqrt3}2`