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Question:

If the tangent at a point P, with parameter t, on the curve x = 4t² + 3, y = 8t³ meets the curve again at a point Q, then the coordinates of Q are:

(t²+3, -t³)

(16t³+3, -4t³)

(t²+3, t³)

(4t²+3, -t³)

Solution:

dx/dt = 8t and dy/dt = 24t²
dy/dx = 3t
y - y₁ = m(x - x₁)
y - (8t³) = 3t(x - (4t² + 3))
Equation is y = 3tx - 8t³
We can substitute and check that option A satisfies the equation.