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Mars2501 [29]
3 years ago
6

Consider the level surface given by

Mathematics
1 answer:
NeTakaya3 years ago
8 0
x^2-y^2+z^2=2

If y=2, then you get x^2-2^2+z^2=2\iff x^2+z^2=6 which is a circle in the x-z plane with radius \sqrt6.

If x=1, then you get 1^2-y^2+z^2=2\iff z^2-y^2=1 which is a hyperbola in the y-z plane.

If y=0, you get another circle in the x-z plane defined by x^2+z^2=2, this time with radius \sqrt2.

If x=2, then 2^2-y^2+z^2=2\iffy^2-z^2=2, another hyperbola in the y-z plane with branches perpendicular to the case where x=1.
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[ Refer to the attachment for steps ]

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D)

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