Answer to Question #133022 in Analytic Geometry for Promise Omiponle

Question #133022
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?
1
Expert's answer
2020-09-14T19:20:27-0400

a) x2+y2z2=1x^2+y^2-z^2=1


x=k=>k2+y2z2=1=>y2z2=1k2x=k=> k^2+y^2-z^2=1=>y^2-z^2=1-k^2

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

If k=±1,y2z2=(yz)(y+z)=0,k=\pm 1, y^2-z^2=(y-z)(y+z)=0, so it is a union of two lines.



y=k=>x2+k2z2=1=>x2z2=1k2y=k=> x^2+k^2-z^2=1=>x^2-z^2=1-k^2

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

If k=±1,x2z2=(xz)(x+z)=0,k=\pm 1, x^2-z^2=(x-z)(x+z)=0, so it is a union of two lines.



z=k=>x2+y2k2=1=>x2+y2=1+k2z=k=> x^2+y^2-k^2=1=>x^2+y^2=1+k^2

The trace is a circle whose radius is 1+k2.\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=kz=k is

never empty. This implies the hyperboloid is connected.


b) The role of yy and zz are interchanged. So now the axis of given hyperboloid

is yy -axis.


c)


x2+y2+2yz2=0=>x^2+y^2+2y-z^2=0=>

x2+(y2+2y+1)z2=1=>x^2+(y^2+2y+1)-z^2=1=>

x2+(y+1)2z2=1x^2+(y+1)^2-z^2=1

Thus it is a translation of the hyperboloid x2+y2z2=1x^2+y^2-z^2=1 by (0,1,0).(0,-1,0).



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