Question

Find the equation of the curve of intersection:
z = x^2 + y^2
4x^2 + y^2 + z^2 = 9

Answer 1

Mostly just need someone who can check my work. I started by setting the first equation equal to 9, so\[x^2+y^2-z=9\]and then set solved both for zero and set them equal to each other \[x^2+y^2-z-9=4x^2+y^2+z^2\]solving I got \[4x^2-x^2+y^2-y^2+z^2-z-9+9=0\]\[3x^2+z^2-z=0\]

Answer 2

Though, I don't think that can be right, because the full question is: Use the gradietn vector to find an equation for th tangent line to the curve of intersection of z = x^2 + y^2 and 4x^2 + y^2 + z^2 = 9 at point P(-1, 1, 2). (it's just find the curve of intersection that I am having difficulty with.)

Answer 3

z = x^2 + y^2

let x^2 = z - y^2

4x^2 + y^2 + z^2 = 9

4(z - y^2) + y^2 + z^2 = 9

4z - 4y^2 + y^2 + z^2 = 9

z^2 + 4z - 3y^2 = 9

(z^2 + 4z + 4) -4 - 3y^2 = 9

(z+2)^2 - 3y^2 = 13

\[\frac{(z+2)^2}{13} - \frac{y^2}{13/3} = 1\]

Answer 4

Thank you! Though, is it really necessary to divide by 13 at the end there? (i.e. wouldn't it be just the same to leave it as \[(z+2)^2-3y^2=13\]?