State the type of the quadratic surface: x2+(y6)2+z2=1 1. hyperboloid of two sheets 2. hyperboloid of one sheet 3. ellipsoid 4. none of these

solution:

quadratic surface,

x^{2}+z^{2}+\frac{y^{2}}{36}=1

=x^{2}+z^{2}=1-\frac{y^{2}}{36}

=\frac{x^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1-\frac{y^{2}}{b^{2}}

this is a hyperbolid of one that intersection at  

y=0

the quadratic surface becomes

x^{2}+z^{2}=1

this is the equation of the unit circle

so,the surface is three circle

-2 is the right

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.