please refer to photo below: By setting one variable constant, find a plane that

Question

please refer to photo below:

By setting one variable constant, find a plane that intersects the graph of z = 2y2 - 4x2 + 2 in a. Parabola opening upward: the plane = (Give your answer by specifying the variable in the first answer blank and a value for it in the second.) Parabola opening downward: the plane = (Give your answer by specifying the variable in the first answer blank and a value for it in the second.) Pair of intersecting straight lines: die plane = (Give your answer by specifying the variable in the first answer blank and a value for it in the second.)

Explanation / Answer

a) x= 0

Then, z = 2y^2 + 2 opens upward

b) y = 0

Then, z = -4x^2 + 2 opens downward

c) z = 2

Then, 2 = 2y^2 - 4x^2 + 2, so

0 = 2y^2 - 4x^2, or

2y^2 = 4x^2, or

y^2 = 2x^2

y = plus or minus sqrt(2) x

These are two lines, both going through (0, 0, 2)

The lines may be written (x, sqrt(2)x, 2) and (x. - sqrt(2) x, 2)

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