I need to find a specific Hyperboloid

**GIPC** · November 5th 2010, 01:42 PM

I have the following question and i'm lost:

Find a Hyperboloid of one sheet ( x^2/a^2 + y^2/b^2 - z^2/c^2 =1 ) that includes the line which connects the dots (0,1,1) and (1,0,-1)

**CaptainBlack** · November 5th 2010, 10:39 PM

Originally Posted by GIPC

I have the following question and i'm lost:

Find a Hyperboloid of one sheet ( x^2/a^2 + y^2/b^2 - z^2/c^2 =1 ) that includes the line which connects the dots (0,1,1) and (1,0,-1)

The line joining the points is:

$X(\lambda)=\lambda (0,1,1)+(1-\lambda)(1,0,-1)$

Now substitute

$x=(1-\lambda)$
$y=\lambda$
$z=2\lambda-1$

into the equation for the hyperboloid and simplify.

Now when you have done that and if you are still having problems come back and ask for more help.

CB

**GIPC** · November 6th 2010, 02:14 AM

I'm sorry but I don't quite follow your logic.

Isn't the line joining the points should be (x-x1/x2-x1)=(y-y1/y2-y1)=(z-z1/z2-z1) ?

and say I substitute into the hyperboloid the x,y,z... all I get is a mess, I still need to find the parameters a,b,c and on the of that the Lambda. And it looks very ugly

**CaptainBlack** · November 6th 2010, 06:56 AM

Originally Posted by GIPC

I'm sorry but I don't quite follow your logic.

Isn't the line joining the points should be (x-x1/x2-x1)=(y-y1/y2-y1)=(z-z1/z2-z1) ?

and say I substitute into the hyperboloid the x,y,z... all I get is a mess, I still need to find the parameters a,b,c and on the of that the Lambda. And it looks very ugly

Which is why I gave you a different form of the line joining the points. Use that and you don't get a mess.

CB

**GIPC** · November 6th 2010, 07:34 AM

I'm sorry but I don't follow. How is this not a mess?
(1-lambda)^2/a^2 + lambda^2/b^2 - (2*lambda -1)^2/c^2 = 1

it gets very ugly very fast.

**CaptainBlack** · November 6th 2010, 09:12 AM

Originally Posted by GIPC

I'm sorry but I don't follow. How is this not a mess?
(1-lambda)^2/a^2 + lambda^2/b^2 - (2*lambda -1)^2/c^2 = 1

it gets very ugly very fast.

It is a quadratic in $$$ \lambda$ , expand it and equate the coefficient of $$$ \lambda^2$ to zero, and the coefficient of $$$ \lambda$ to zero and the constant term to one.

That gives you a system of three simultaneous equations in $1/a^2$ , $1/b^2$ and $1/c^2$ that can be solved by inspection.

CB

**GIPC** · November 6th 2010, 10:45 AM

Well, that's what I get:
L^2*(1/a^2 + 1/b^2 -4/c^2) + L*(-2/a^2 -4/c^2) +1/a^2 + 1/c^2 = 1
so I get the system
1/a^2 + 1/b^2 -4/c^2=0,
-2/a^2-4/c^2=0,
1=1/a^2 +1/c^2

In order to solve I have to use imaginary numbers, it seems...
where did I go wrong?

**GIPC** · November 6th 2010, 11:12 AM

Well, what can it be? The Hyperboloid exists, it is defined in my HW assignment, I just need to find it

I checked carefully for typos, and can't seem to track any.

Is there maybe another way to approach the problem? Say if i substitute the 2 given points in the generic equation, I get 2 equations, and I only need to find one more- how can I find something else to give me another equation?

**CaptainBlack** · November 6th 2010, 11:29 AM

Originally Posted by GIPC

Well, what can it be? The Hyperboloid exists, it is defined in my HW assignment, I just need to find it

I checked carefully for typos, and can't seem to track any.

Is there maybe another way to approach the problem? Say if i substitute the 2 given points in the generic equation, I get 2 equations, and I only need to find one more- how can I find something else to give me another equation?

Hang on a minute, you will notice I have deleted the previous post temporarily as there is a mistake in the algebra (I used your quadratic rather than rederiving it), which I can now correct:

You should have:

$\dfrac{(1-\lambda)^2}{a^2}+\dfrac{\lambda^2}{b^2}-\dfrac{(2\lambda-1)^2}{c^2}=1$

or:

$\lambda^2\left( \frac{1}{a^2}-\frac{1}{b^2}-\frac{4}{c^2}\right) + \lambda \left(-\frac{2}{a^2}+\frac{4}{c^2} \right) +\left( \frac{1}{a^2} - \frac{1}{c^2}\right)=1$

so our simultaneous equations are:

$\frac{1}{a^2}-\frac{1}{c^2}=1$

$\frac{1}{a^2}-\frac{2}{c^2}=0$

$\frac{1}{a^2}+\frac{1}{b^2}-\frac{4}{c^2}=0$

Which have solution $\frac{1}{a^2}=2,\ \frac{1}{b^2}=2,\ \frac{1}{c^2}=1$

CB

**GIPC** · November 6th 2010, 11:42 AM

Thanks a bunch CB

If it's not any trouble, can you explain (or link to an explanation) about the initial step when you found the line joining the points using X(lambda)=LAMBDA*p1 + (1-LAMBDA)*p2? where did this come from? We never discussed such forms in our classes.

**CaptainBlack** · November 6th 2010, 03:16 PM

Originally Posted by GIPC

Thanks a bunch CB

If it's not any trouble, can you explain (or link to an explanation) about the initial step when you found the line joining the points using X(lambda)=LAMBDA*p1 + (1-LAMBDA)*p2? where did this come from? We never discussed such forms in our classes.

It is the vector form of the line joining the two points, it obviously depends on only one parameter and when $$$ \lambda=1$ it is $$$ p1$ and when $$$ \lambda=0$ it is $$$ p2$ .

It is simply $$$ p1$ plus a multiple of the vector from $$$ p1$ to $$$ p2$

CB

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