(a) surely you can just plug in a value for t!
x = 2cos(8π/3) = 2(-1/2) = -1
y = 2sin(8π/3) = 2(√3/2) = √3
(b) Now you have a point and a slope, so the tangent line is
y-√3 = 1/√3 (x+1)
A circle is defined by the parametric equations
x = 2cos(2t) and y = 2sin(2t) for t is all real numbers.
(a) Find the coordinates of the point P on the circle when t = (4*pi)/3.
(b) Find the equation of the tangent to the circle at P.
Can you please show your working to help me understand how you reach your answer? Thanks - Jichu
1 answer
1 answer