Answer
$y = - \sqrt {15}(x+ 2)$
and
$y = \sqrt {15}(x+ 2)$
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Work Step by Step
1. Identify the point where the curve cross it self.
According to the graph: $(-2,0)$
2. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{d(sin(t) + 2sin(2t))}{dt} = cos(t) + 4cos(2t)$
$\frac{dx}{dt} = \frac{d(cos(t) + 2cos(2t))}{dt} = -sin(t) - 4sin(2t)$
Thus, $\frac{dy}{dx} = \frac{cos(t) + 4cos(2t)}{-sin(t) - 4sin(2t)}$
3. Identify both $t$ values at (-2,0):
$-2 = cos(t) + 2cos(2t)$
$0 = sin(t) + 2sin(2t)$
$0 = sin(t) + 2(2sin(t)cos(t))$
$0 = sin(t) (1 + 4cos(t))$
$sin(t) = 0$; $t = 0 \space or \space t = \pi$
Check if $x$ = -2
$x = cos(0) + 2cos(2*0) = 1 + 2*1 = 3 \neq -2$
$x = cos(\pi) + 2cos(2\pi) = -1 + 2*(1) = 1 \neq -2$
$1 + 4cos(t) = 0$
$cos(t) = - \frac{1}{4}$
$t =± arccos(-\frac{1}{4})$
Check if $x$ = -2:
$x = cos(arccos(-\frac{1}{4})) +2cos(2*(arccos(-\frac{1}{4}))) = -2$
$x = cos(-arccos(-\frac{1}{4})) +2cos(2*(-arccos(-\frac{1}{4}))) = -2$
4. Calculate $\frac{dy}{dx}$ for these values of $t$:
$\frac{dy}{dx} = \frac{cos(arccos(-\frac{1}{4})) + 4cos(2(arccos(-\frac{1}{4})))}{-sin(arccos(-\frac{1}{4}))-4sin(2(arccos(-\frac{1}{4})))} \approx -3.873 \approx - \sqrt {15}$
$\frac{dy}{dx} = \frac{cos(-arccos(-\frac{1}{4})) + 4cos(2(-arccos(-\frac{1}{4})))}{-sin(-arccos(-\frac{1}{4}))-4sin(2(-arccos(-\frac{1}{4})))} \approx 3.873 \approx \sqrt {15}$
5. Write the equations:
$y = - \sqrt {15}(x+ 2)$
and
$y = \sqrt {15}(x+ 2)$
