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

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92

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