Curve of intersection of a plane and curve

[Yitin]

Homework Statement

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...

 

[LCKurtz]

Homework Statement

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...

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.

 

[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)

 

[LCKurtz]

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)

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