A curve is given parametrically by the equation x = 3-4sinT and y = 4+3cosT for 0 <= T <= 2pi. What are all the points (x,y) at which the curve has a vertical tangent?
dx/dt=-4cost
dy/dt=-3sint
so dy/dx=dy/dt * dt/dx
dy/dx=-3sin(t)/(-4cos(t))
So it's infinite when cos(t)=0, with t between 0 and 2pi
cos(pi/2)=cos(3pi/2)=0
So at t=pi/2 and t=3pi/2, there is a vertical tangent.
t=pi/2 -> x = 3-4sin (pi/2)=3-4=-1, y = 4+3cos(pi/2)=4 ------> (-1,4)
t=3pi/2 -> x = 3-4sin (3pi/2)=3-4*(-1)=7, y = 4+3cos(3pi/2) = 4 ------> (7,4)