# Curve of intersection of a plane and curve

1. Jul 14, 2012

### Yitin

1. The problem statement, all variables and given/known data
I have to find the slope of the tangent line at (-1,1,5) to the curve of intersection of the surface z = x2 + 4y2 and the plane x = -1

2. The attempt at a solution
I really am having trouble figuring out where to start. The question is even numbered, and the only one like it in the chapter, so I have been having trouble figuring out what to do.
I tried googling it different ways, the explanation of a problem that is kinda like this one told me to take the gradient
2xi + 8yj - k
and then plug in (-1,1,5) and that would be the normal n1
And I was supposed to do the cross of n1 and the normal of the plane, and then some more stuff but their problem was different, and I am not really sure where to go from here, or if this is even right...

2. Jul 14, 2012

### LCKurtz

You wouldn't usually use the term "slope" for the tangent to a 3D curve. You would ask for a direction vector. If you plug in $x=-1$ you get $z=1+4y^2$. With these you can parameterize the curve $\vec R = \langle -1,y,1+4y^2\rangle$. Now it should be easy to find the direction vector for the tangent to the curve.

3. Jul 15, 2012

### Yitin

Thank you, that helped greatly, though I am still not the best at getting that curve.

A later problem has z=√(x2 + 4y2) and 3z = x + 2y + 8
I was trying to make the two Zs equal to each other, and solve for x or y, but I couldn't get any of them separate.
Same thing, finding the tangent line at a point (3,2,5)

4. Jul 15, 2012

### LCKurtz

Forum policy is that you start a new thread if you have a new problem. You are more likely to get an answer.