# Thread: Parametrize curve of intersection

1. ## Parametrize curve of intersection

The problem is:
"Parametrize the intersection of the hyperboloid $\displaystyle x^2+y^2=z^2+1$ and the plane $\displaystyle z=ay$, where $\displaystyle a$ is a positive real number. Give the range of the values taken by the parameter. -Hint: Consider the cases 0<a<1 and a > 1 separately, and remember that x, y and z are all real."

So here's what I did:
I first plugged in the equation of the plane into the equation of the hyperboloid: $\displaystyle x^2+(1-a^2)y^2=1$

If a<1, I get an ellipse of the form: $\displaystyle \frac {x^2}{1^2} +\frac{y^2}{\frac{1}{(\sqrt{1-a^2})^2}}=1$, similar to $\displaystyle \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$
From an online source, I found that $\displaystyle x=\cos\phi$ and $\displaystyle y=\frac{1}{\sqrt{1-a^2}}\sin\phi$, and therefore $\displaystyle z=\frac{a}{\sqrt{1-a^2}}\sin\phi$ and that $\displaystyle 0\leq \phi \leq 2\pi$

If a>1, the equation can be put in the form $\displaystyle \frac {x^2}{1^2} -\frac{y^2}{\frac{1}{(\sqrt{a^2-1})^2}}=1$ which is the equation of a hyperbola. Once again, from an online source (Wolfram MathWorld), I found that I could represent both branches of the hyperbola with $\displaystyle x=\sec\phi$, $\displaystyle y=\frac{1}{\sqrt{a^2-1}}\tan\phi$ and $\displaystyle z=\frac{a}{\sqrt{a^2-1}}\tan\phi$, where $\displaystyle -\pi < \phi < \pi$ with discontinuities at $\displaystyle \pm \frac{\pi}{2}$

I've seriously spent a lot of time on this problem, and this is the best I can come up with. Can someone tell me if it's acceptable? If not, what should I do?

2. It looks good to me. Of course, I can't speak for your teacher!

3. k

4. I don't see why you have to do all that. If I substitute the plane into the hyperboloid, I get for $\displaystyle a\ne 1$:

$\displaystyle x^2+y^2=a^2 y^2+1$ so that $\displaystyle y=\pm \sqrt{\frac{1-x^2}{1-a^2}}$ and $\displaystyle z=ay$. So a parametric representation of the intersection is:

$\displaystyle \left\{x,\pm \sqrt{\frac{1-x^2}{1-a^2}},\pm a\sqrt{\frac{1-x^2}{1-a^2}}\right\}$

with $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$

and:

$\displaystyle |x|>a$ for $\displaystyle a>1$

And I think for the case $\displaystyle a=1$, then $\displaystyle z=y$ which then means $\displaystyle x^2=1$ so that the paramaterization is $\displaystyle \{\pm1,y,y\}$.

The yellow contours below is this paramaterization for a=1/2 and a=2

5. Originally Posted by shawsend
I don't see why you have to do all that. If I substitute the plane into the hyperboloid, I get for $\displaystyle a\ne 1$:

$\displaystyle x^2+y^2=a^2 y^2+1$ so that $\displaystyle y=\pm \sqrt{\frac{1-x^2}{1-a^2}}$ and $\displaystyle z=ay$. So a parametric representation of the intersection is:

$\displaystyle \left\{x,\pm \sqrt{\frac{1-x^2}{1-a^2}},\pm a\sqrt{\frac{1-x^2}{1-a^2}}\right\}$

with $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$

and:

$\displaystyle |x|>a$ for $\displaystyle a>1$

And I think for the case $\displaystyle a=1$, then $\displaystyle z=y$ which then means $\displaystyle x^2=1$ so that the paramaterization is $\displaystyle \{\pm1,y,y\}$.

The yellow contours below is this paramaterization for a=1/2 and a=2
But how do you know that $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$? I can't see that...

That thumbnail was really helpful! Did you use Maple to draw the surface and the intersection?

6. Ok, I think that should be $\displaystyle |x|<1$ when $\displaystyle a<1$. Also then $\displaystyle |x|>1$ for $\displaystyle a>1$. Also, I drew that in Mathematica using this code for the a=1/2 one:

Code:
a = 1/2;
p1 = ParametricPlot3D[{x, Sqrt[1/(1 - a^2) - x^2/(1 - a^2)],
a*Sqrt[1/(1 - a^2) - x^2/(1 - a^2)]}, {x, -1, 1},
PlotStyle -> {Thickness[0.008], Yellow}]
p2 = ParametricPlot3D[{x, -Sqrt[1/(1 - a^2) - x^2/(1 - a^2)],
(-a)*Sqrt[1/(1 - a^2) - x^2/(1 - a^2)]}, {x, -1, 1},
PlotStyle -> {Thickness[0.008], Yellow}]
pic1 = Show[{ContourPlot3D[{x^2 + y^2 == z^2 + 1, z == (1/2)*y}, {x, -3, 3},
{y, -3, 3}, {z, -3, 3}], p1, p2}]

7. Thank you so much!