Answer
The points of horizontal tangency are $(0,3)$ and $(0,-3)$
The points of vertical tangency are $(3,0)$ and $(-3,0)$
Work Step by Step
The point of horizontal tangency will occur when $dy/dx$=0, therefore when dy is zero, in this case:
$y=3sin(θ)$
$dy=(3cos(θ))dt$
dt can't be equal to 0, so: $3cos(θ) = 0$
$θ = \frac{\pi}{2}$ or $\theta = \frac{3\pi}{2}$
Substitute the values at $\theta$=$\frac{\pi}{2}$:
$x=3cos(θ) = 3cos(\frac{\pi}{2})=3*0 = 0$
$y=3sin(θ) = 3sin(\frac{\pi}{2}) = 3 *1 = 3$
Therefore, one point of horizontal tangency is (0,3)
Substitute the values at $θ = 3\pi/2$:
$x=3cos(θ) = 3cos(3\pi/2) = 3*0 = 0$
$y=3sin(θ) = 3sin(3\pi/2) = 3*-1 = -3$
Therefore, the other point of horizontal tangency is (0,-3)
------
The point of vertical tangency will occur when dx is equal to 0:
$x = 3cos(θ)$
$dx = (-3sin(θ))dt$
dt can't be equal to 0, so: $-3sin(θ) = 0$
$sin(θ) = 0$
$θ = 0$ or $θ = \pi$
Solve for "x" and "y" when $θ = 0$
$x = 3cos(θ) = 3cos(0) = 3 * 1 = 3$
$y = 3sin(θ) = 3sin(0) = 3 *0 = 0$
Solve for "x" and "y" when (θ = $\pi$)
$x = 3cos(\pi) = 3*-1 = -3$
$y = 3sin(\pi) = 3*0 = 0$
Therefore, the points of vertical tangency are $(3,0)$ and $(-3,0)$
