Here is the problem exactly how it is written on my paper...

Consider the surfaces x^2+2y^2-z^2+3x=1 and 2x^2+4y^2-2z^2-5y=0.
a. What is the name of each surface?
b. Find an equation for the plane which contains the intersection of these two surfaces.

That is the question. I got part "a" and I am pretty sure it is right. What I really have no idea how to do is "b". Can someone tell me the answer to this and please explain to me how to do it? Thanks in advance.

2. anyone know the answer to this?

3. So you have:

$\displaystyle x^2+2y^2-z^2+3x=1$

$\displaystyle 2x^2+y^2-2z^2-5y=0$

They intersect when the z-values are the same. So solve for z for both. Say you get $\displaystyle z_1=\pm f(x,y)$ and $\displaystyle z_2=\pm g(x,y)$. Now equate them:

$\displaystyle f_n(x,y)=g_n(x,y)$

and solve for y. I get for one, a linear equation in y:$\displaystyle y_1=mx+b$. Alright, then the intersection of the two quartic surfaces are the combination of all these expressions. For example, I get for one:

$\displaystyle y_1(x)=-2/5(-1+3x)$ and $\displaystyle f_1(x,y)=\sqrt{-1+3x+x^2+2y^2}$

and so in parametric form, that part of the intersection would be $\displaystyle \{x,y_1(x),f_1(x,y_1(x))\}$. The plot below shows the two surfaces, and the yellow trace is $\displaystyle \{x,y_1(x),f_1(x,y_1(x))\}$

4. To get the equation of the plane we can find 3 points on the plane

and then proceed as I'm sure you know how to do.

1)

2)

Multiply 1) by -2 and add to 2) -3y^2 -6 x - 5 y = -2

or 3y^2 + 6x + 5y = 2
If y = 0 x= 1/3 and z = + 1/3

if x = 0 y= 1/3 or -2 let's use -2 then z= +sqrt(7)

For the 3 pts use (1/3,0,+1/3) and (0,-2,sqrt(7)

5. Here's a plot of the plane. I used the equation:

$\displaystyle a_1(x-x_1)+a_2(y-y_1)+a_3(z-z_1)=0$ where $\displaystyle \{a_1,a_2,a_3\}$ is a normal vector to the plane. That means I had to use four points along the yellow curves above to calculate the normal vector (three equations in three unknowns and one point for the (x,y,z) values as well. Probably an easier way though. This is tough for me.

6. Originally Posted by Calculus26
To get the equation of the plane we can find 3 points on the plane

and then proceed as I'm sure you know how to do.

1)

2)

Multiply 1) by -2 and add to 2) -3y^2 -6 x - 5 y = -2

or 3y^2 + 6x + 5y = 2
If y = 0 x= 1/3 and z = + 1/3

if x = 0 y= 1/3 or -2 let's use -2 then z= +sqrt(7)

For the 3 pts use (1/3,0,+1/3) and (0,-2,sqrt(7)
So you're saying I should use the points (1/3,0,1/3), (1/3,0,-1/3), and (0,-2,sqrt(7)) as the three points, right? Just want to make sure, because when I used them the normal vector's k component came out to be 0. Just seemed a little weird to me, so I just wanted to make sure I understood you correctly.

Thanks for the help and info everyone.

7. ^^??

8. ??

9. If the k component is 0 that simply means you have a vertical plane.

Not out of the realm of possibility