Quadratic surfaces standard form help

[Mark53]

Homework Statement

[/B]
Suppose a quadratic equation in 3 variables is put into a standard form represents a hyperboloid of one sheet. This hyperboloid has the property that:

• the cross section through z= 0 is a circle of radius 1;

• the cross section through x= 1 is the two straight lines given (in the plane x = 1) by y = ± (1/2) z

What is the standard form and the original equation in quadratic form?

Homework Equations

[/B]
hyperboloid of one sheet is given by

x^2/a^2+y^2/b^2-z^2/c^2=1

The Attempt at a Solution

[/B]
using z=0

x^2/a^2+y^2/b^2=1 which is an ellipse

Given that it is a circle it means that b=a which means that it equals 1 as that is what the radius is.

x^2+y^2-z^2/c^2=1

How would I find C from here

 

[haruspex]

How would I find C from here

Use the other piece of information:

the cross section through x= 1 is the two straight lines given (in the plane x = 1) by y = ± (1/2) z

 

[Mark53]

Use the other piece of information:

1+(1/4)z^2-z^2/c^2=1

(1/4)z^2-z^2/c^2=0

z^2((1/4)-c^2)=0

(1/4)-c^2)=0

c^2=1/4

c=1/2

therefore

x^2+y^2-z^2/(1/2)^2=1

 

[pasmith]

Why not simplify 1/(1/2)^2 as 4?

 

[haruspex]

$(1/4)z^2-z^2/c^2=0$

$z^2((1/4)-c^2)=0$

Check that step.

 

[Mark53]

Check that step.

that would mean that c=2

 

[haruspex]

that would mean that c=2

Yes.