Show that the curve has two tangents find their equations

Question

Show that the curve x=2cos(t), y= 7sin(t)cos(t) has two tangents at (0,0) and find their equations.

y=_____ (Smaller slope)

y= _____ (Larger slope)

Answer 1

If x = 2cos(t) then x/2 = cos(t) and t = cos^-1(x/2) meaning we can draw this triangle (since cos(t) = adjacent/hypotenuse = x/2):

We can solve for the other side using the Pythagorean Theorem making the opposite side is √(2² - x²) = √(4 - x²)

Then sin(t) = opposite/hypotenuse = √(4 - x²)/2

Using y = 7sin(t)cos(t), we can substitute the above and say:

y = 7(√(4 - x²)/2)(x/2)

y = (7/4)x√(4 - x²)

y' (using the product rule) is

y' = 7/4√(4 - x²) + (7/4)x(1/2(4 - x²)^-1/2(2x)

y' = 7/4√(4 - x²) + (7x²)/(4√(4 - x²))

When x = 0, y'(0) = 7/4√(4 - 0²) + (7•0²)/(4√(4 - 0²)) = (7/4)√4 + 0 = 7√4/4 HOWEVER, we must remember that √4 is either +2 or -2 so y' = 7(2)/4 and 7(-2)/4 = 7/2 and -7/2

So the equations (since the y-intercept is zero) are

y = -7/2x

y = 7/2x

Answer 2

dy/dx = (dy/dt)/(dx/dt) = (7(cos²t - sin²t))/(-2sint)= - 7cos2t/(2sint). At t = π/2 x(π/2) = y(π/2) = 0 and slope of tangent line dy/dx = 7/2. Then at t = 3π/2 x(3π/2) = y(3π/2) = 0 and slope of tangent line is dy/dx = - 7/2

y = 7/2x and y = - 7/2x equations of this tangent lines.