well, we have
sin(2t) = x/4
cos(2t) = y/3
so
(x/4)^2 + (y/3)^2 = 1
x^2/16 + y^2/9 = 1
9x^2 + 16y^2 = 144
dy/dx = (dy/dt)/(dx/dt) = -6sin2t / 8cos2t = -3/4 tan2t
so at t = π/6, we have dy/dx = -3/4 * √3 at the point (2√3,3/2)
so the tangent line is
y - 3/2 = -3√3/4 (x - 2√3)
see the graphs at (eliding the extra spaces)
www.w olframalpha.c om/input?i=plot+9x%5E2+%2B+16y%5E2+%3D+144%2C+y+-+3%2F2+%3D+-3%E2%88%9A3%2F4+%28x+-+2%E2%88%9A3%29
Find a Cartesian equation relating and corresponding to the parametric equations: x=4sin(2t), y=3cos(2t).
Write your answer in the form P(x,y)=0, where P(x,y) is a polynomial in x and y, such that the coefficient of y^2 is 16.
b)Find the equation of the tangent line to the curve at the point corresponding to t=pi/6
1 answer