Hi all, first time poster but if this goes well will probably be seeking out more help.

So we are doing parametric equations and I understand how to find the slope, however from their I'm totally lost. >.<

Please don't just answer, I have an exam soon and if you could explain why, it would be wonderful.

The original question is:

Find an equation of the tangent line to this curve

x=(t^2)-8t

y=(t^2)+7t

So I know you take the derivative of both sides to find dx/dt and dy/dt

dx/dt=2t-8

dy/dt=2t+7

So the slope is equal to...

(2t+7)/(2t-8)

The T value they give is 8/2 (4?)

So..

2(4)+7/2(4)-8

15/0

Slope is undefined....

From there I have no idea

_______________________

However if the slope were to work, such as with this equation...

x=(t^3)-3t

y=t^2

@ t= 3

I get the slope 1/4

so

y=(x/4)+b

However, I don't know what values to put for x and y!!

Anyway. Thank you for any help.