1. intersection between two surfaces

I have two surfaces: x^2+y^2-z^2=0 (cone) and x^2-y^2-z=0 (hyperbolic paraboloid). I'm asked to find their intersection line in its parametric representation and to give the bounds of t.

I tried to take z=x^2-y^2 from the second equation and to use it in the first one, but it leads to a complicated equation and I'm not sure that this is the correct way to deal with this question.

2. Hint :

Write for the cone:

$\displaystyle x=\left |{z}\right |\cos \theta,\;y=\left |{z}\right |\sin \theta\quad (\theta\in{[0,2\pi]})$

Substituting values into the paraboloid equation:

$\displaystyle z(z\cos 2\theta-1)=0$

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Fernando Revilla

3. thanks for your answer. I thought about representing the surfaces by vectors, and got the same result.

It leads to $\displaystyle (x,y,z) = {1 \over {\cos 2\theta }}\left( {\cos \theta ,\sin \theta ,1} \right)$ and I don't understand why this is a (straight) line. $\displaystyle \theta$ is limited to $\displaystyle [0,\pi /4)$ for this curve.

4. Originally Posted by fict
...and I don't understand why this is a (straight) line.
Why do you say that is a straight line?

$\displaystyle \theta$ is limited to $\displaystyle [0,\pi /4)$ for this curve.
You need to add more values for $\displaystyle \theta$. Take into account that if $\displaystyle (x,y,z)$ belongs to the line, then:

$\displaystyle (-x,y,z),\;(x,-y,z),\;(-x,-y,z)$

also belong to the line.

Regards.

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Fernando Revilla

5. Why do you say that is a straight line?
In the question they asked about a line, does it mean that they refer to some general curve? not just a straight line?