Answer to Question #133022 in Analytic Geometry for Promise Omiponle

Question

a.) Find and identify the traces of the quadric surfacex^2+y^2-z^2= 1 and use the traces to classify the graph of the surface.
b.) If we change the equation in part a.) to x^2-y^2+z^2= 1, how is the graph affected?
c.) What if we change the equation in part a.) to x^2+y^2+ 2y-z^2= 0? How is the graph affected in this case?

Answer 1

a) $x^2+y^2-z^2=1$

$x=k=> k^2+y^2-z^2=1=>y^2-z^2=1-k^2$

The trace is a hyperbola when $k\not=\pm 1.$

If $k=\pm 1, y^2-z^2=(y-z)(y+z)=0,$ so it is a union of two lines.

$y=k=> x^2+k^2-z^2=1=>x^2-z^2=1-k^2$

The trace is a hyperbola when $k\not=\pm 1.$

If $k=\pm 1, x^2-z^2=(x-z)(x+z)=0,$ so it is a union of two lines.

$z=k=> x^2+y^2-k^2=1=>x^2+y^2=1+k^2$

The trace is a circle whose radius is $\sqrt{1+k^2}.$

Therefore the surface is a stack of circles, whose traces of other directions are

hyperbolas. So it is a hyperboloid. The intersection with the plane $z=k$ is

never empty. This implies the hyperboloid is connected.

b) The role of $y$ and $z$ are interchanged. So now the axis of given hyperboloid

is $y$ -axis.

c)

$x^2+y^2+2y-z^2=0=>$

$x^2+(y^2+2y+1)-z^2=1=>$

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

Thus it is a translation of the hyperboloid $x^2+y^2-z^2=1$ by $(0,-1,0).$