# Math Help - Intersection of two surfaces

1. ## Intersection of two surfaces

Note: This is from a Vector Calculus course, so I hope it belongs in this section.

The question:
For what values of 'a' do the surfaces $x^2 + y^2 + (z - 1)^2 = 1$ and $a(z+1)^2 = x^2 + y^2, (z \ge -1)$ not intersect?

My attempt:
I rearranged the first equation like so:
$x^2 + y^2 = 1 - (z - 1)^2$

Then equated it with the other:
$a(z+1)^2 = 1 - (z - 1)^2$

Usually I'd solve for z to find the conditions for the intersection, but we have two variables now.

I then solved for a:
$a = \frac{1 - (z - 1)^2}{(z+1)^2}$

So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.

I don't think I'm doing this correctly. Any assistance would be great!

2. $a(z+1)^2 = 1 - (z - 1)^2$

$(a+1)z^2+2(a-1)z+a=0$

This equation has no solution when the discriminant is negative

3. Ahh that makes sense, thank you!

4. Originally Posted by Glitch
Note: This is from a Vector Calculus course, so I hope it belongs in this section.

The question:
For what values of 'a' do the surfaces $x^2 + y^2 + (z - 1)^2 = 1$ and $a(z+1)^2 = x^2 + y^2, (z \ge -1)$ not intersect?

My attempt:
I rearranged the first equation like so:
$x^2 + y^2 = 1 - (z - 1)^2$

Then equated it with the other:
$a(z+1)^2 = 1 - (z - 1)^2$

Usually I'd solve for z to find the conditions for the intersection, but we have two variables now.

I then solved for a:
$a = \frac{1 - (z - 1)^2}{(z+1)^2}$
This is the "wrong way". Solve the equation for z, not a. That is a quadratic equation for z so there will be no real number solution for z if the "discriminant" ( $b^2- 4ac$) is negative.

So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.

I don't think I'm doing this correctly. Any assistance would be great!