here, im using the method for finding the vector equation

**given the plane 2x + 6y + z = 6 intersects the paraboloid z = x^2 + y^2, find and name this curve of intersection.**

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so far, i rearranged the plane equation to a function of z and let it equal te equation of the paraboloid.

**this gave me:**

x^2 + y^2 = 6 -2x - 6y

6 = x^2 + 2x + y^2 + 6y

6 = [x^2 + 2x + 1^2] + [y^2 + 6y +3^2] -10

16 = [x + 1]^2 + [y + 3]^2

**4^2 = [x + 1]^2 + [y + 3]^2**

this then gave me the projection downwards of the function, a circle. but the function should be an ellipse. (which cant be called a function y of x, because it fails the vertical test, not to clear on this)

to get the function of the curve of intersection, (here is where im not sure about where im going) i wanted x infunction of t and y in function of t (for parametric equations)

x = f(t)

y = g(t)

**so i went:**

*rule:*(cos(t))^2 + (sin(t))^2 = 1

[(x +1)/4]^2 + [(y + 3)/4]^2 = 1

(x +1)/4 = cos(t) => x + 1 = 4*cos(t) => x = 4*cos(t) - 1

(y + 3)/4 = sin(t) => y + 3 = 4*sin(t) => y = 4*sin(t) - 3

[(respectively):

*eqn, first eqn, second eqn*]

*[where o<= t<= 2*pi]*

**from,**

4^2 = [x + 1]^2 + [y + 3]^2

=> 4^2 = [4*cos(t)]^2 + [4*sin(t)]^2

4^2 = 16*[cos(t)]^2 + 16*[sin(t)]^2

4^2 = 16*([cos(t)]^2 + [sin(t)]^2)

**4^2 = 16*([cos(t)]^2 + [sin(t)]^2)**

**substituting**

*second eqn x*and y values into the plane:z = 6 - 2x - 6y

=> z = 6 - 2*[4*cos(t) -1] - 6*[4*sin(t) - 3]

z = 6 - 8*cos(t) +2 - 24*sin(t) + 18

**z = -8*cos(t) - 24*sin(t) + 26**

this is what i have done so far (stop me if im doing something completely wrong please ^^') but i'm a little lost in what the text book is telling me

**therefore the parametric equations for the equation**

*here is where everything goes iffy (if not already somewhere above).*

when i use the parametric equation, because the projection is a circle the parametric equations would be:

when i use the parametric equation, because the projection is a circle the parametric equations would be:

x = 4*cos(t) - 1

y = 4*sin(t) - 3

z = -8*cos(t) - 24*sin(t) + 26

*[where o<= t<= 2*pi]*

so the corresponding vector equation is:

r(t) = [4*cos(t) - 1]

**i**+ [4*sin(t) - 3]

**j**+ [-8*cos(t) - 24*sin(t) + 26]

**k**

*[where o<= t<= 2*pi]*

sorry if my explaination is unclear in some parts. im not sure if this method gets me to the answer i want, but is there possibly another way to solve this problem? this was the only way i could think of.

the text book didnt exactly teach the topic but there was an example that was similar to this problem im working on so i was working alot of this from that example (which was a really simple example and only had four lines for working).