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 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: 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).