# please help with a differentiation question 10 pts?

The curve C has parametric equations x=t^2+4t and y=t^3+t^2.

The tangent to the curve at the point A where t=2 is denoted by k.

(I) find the cartesian equation of k

(II) the tangent k meets the curve C again at point B. Use a non calculator method to find the coordinates of B.

I just need help with part II of the question.

Thanks!

### 2 Answers

- Ian HLv 78 years agoFavorite Answer
The curve C has parametric equations x = t^2 + 4t and y = t^3 + t^2.

dy/dt = 3t^2 +2t, dx/dt = 2t + 4

dy/dx = (3t^2 +2t)/(2t + 4)

(I) Let cartesian equation of tangent be y = mx + c

when t = 2, y = 12, slope m = dy/dx = 16/8 = 2, x = 12

12 = 2*12 + c, so, c = -12

y = 2x - 12

(II) Substituting parametric y and x values into the line equation,

all points where tangent meets the curve are solutions of

t^3 + t^2 = 2(t^2 + 4t ) - 12

t^3 - t^2 - 8t + 12 = 0

But we already expect the coincident solution t = 2 which helps

(t + 3)(t - 2)^2 = 0

Point B has t = -3, x = -3, y = -18

An alternative but more cumbersome approach is

t^2 + 4t - x = 0 select one of the roots, (both work)

t = √(x + 4) - 2

y = t^2*(t + 1) = [-4√(x + 4)+ x + 8 ][√(x + 4) - 1]

y = (x + 12)√(x + 4) - 5x - 24

The tangent k meets the curve C again when

2x - 12 = (x + 12)√(x + 4) - 5x - 24

(x + 12)√(x + 4) = 7x + 12

(x^2 + 24x + 144)(x + 4) = 49x^2 + 168x + 144

x^3 - 21x^2 + 72x + 432 = 0

We already expect the coincident solution x = 12

(x + 3)(x - 12)^2 = 0

Point B has x = -3, t = -3, y = -18

Regards - Ian

- knakeLv 44 years ago
u can take the divide via t^3 up and the capacity will become t^-3 so the answer is-: (6t - 12t^2) (-3t^-4) =-3(6t - 12t^2)/ t^4 [now take t^-4 down and the capacity is t^4] = (-18t + 36t^2) / t^4