By setting one variable constant, find a plane that intersects the graph of z =2

Question

By setting one variable constant, find a plane that intersects the graph of z=2y^2 %u22126x^2 +1 in a:

(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.)

(b) 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.)

(c) Pair of intersecting straight lines: the plane ( ) = ( )
(Give your answer by specifying the variable in the first answer blank and a value for it in the second.)

need answers in the parenthesis. thanks

By setting one variable constant, find a plane that intersects the graph of z=2y^2 %u22126x^2 +1 in a:

(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.)

(b) 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.)

(c) Pair of intersecting straight lines: the plane ( ) = ( )
(Give your answer by specifying the variable in the first answer blank and a value for it in the second.)

Explanation / Answer

z = 2y^2 -6x^2 +1

if we make either on of yt or x the equation we get will be of a parabola

for general parabolic equation of y = ax^2 + b

opening upward if a>0 and downward if a<o

(a) for opening upward make x = constant =k

(x) =( k), k can be given any constant value 1,2,3 etc.

(b)for opening downward y = constant = k

(y )= (k)

(c)general equation of pair of straight line => ax^2+by^2+2hxy = 0

thus we have to make constant = 0

thus put z = 1 and equation becomes 2y^2 - 6x^2 = 0

thus answer is (z) = (1)

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