Find the equations of the tangents to the given curve that pass through the point (12, 10).
x = 6t^2 + 6
y = 4t^3 + 6
I'm following some example but I get confused after
The answers are:
I understand that m=1 and that makes x-2. But I dont understand is that at one point it looks like m = -2 which is seen in -2x+34 but I dont see how they got m = -2 can someone explain this please.
Oh, wait, I'm sorry. I didn't realize you had already figured that out. Somehow I missed the last paragraph in your original post.
I'm kind of with you now. It's not obvious to me why there are two tangents at that point.
I have a very similar problem in my calculus book.
- Find equations of the tangents to the curve and that pass through the point (4,3).
In this problem I can only come up with y=x-1. So I'm confused now too. It makes sense that parametric curves would have two tangents at certain points since they tend to form loops. I just found a resource that might shed some light on how these work:
If you scroll down to example 1, they discuss this.
Pauls Online Notes : Calculus II - Tangents with Parametric Equations