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## Main Question or Discussion Point

this is the problem and i've worked for mostof it, but im not sure i've got the right way to go about it, or if the ansewr i have found it right.

here, im using the method for finding the vector equation

__________________________________________

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

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

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)

[(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):

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)

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

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

x = 4*cos(t) - 1

y = 4*sin(t) - 3

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

so the corresponding vector equation is:

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

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).

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.**__________________________________________

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).