SOLUTION: For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Algebra ->  Finance -> SOLUTION: For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.       Log On


   

Question 1107273: For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.
Found 2 solutions by Fombitz, greenestamps:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Differentiate implicitly to find the derivative,
dy%2Fdx=%284x-y%29%2F%28x-y%29
Set the derivative equal to zero,
%284x-y%29%2F%28x-y%29=0
4x-y=0
y=4x
So when x=2,
y=4%282%29
y=8
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.
(2,8)
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.
So the tangent line is y=8.
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.
.
.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Use implicit differentiation to find an expression for dy/dx and show that there is a value of y for which the derivative is zero when x is 2.

4x%5E2%2By%5E2=48%2B2xy

8x%2B2y%2A%28dy%2Fdx%29+=+2x%2A%28dy%2Fdx%29%2B2y
2y%2A%28dy%2Fdx%29+-+2x%2A%28dy%2Fdx%29+=+2y-8x
%282y-2x%29%2A%28dy%2Fdx%29+=+2y-8x
dy%2Fdx+=+%282y-8x%29%2F%282y-2x%29+=+%28y-4x%29%2F%28y-x%29

When x=2, the derivative is y-8%29%2F%28y-2%29; it is zero when y is 8.

The tangent to the graph at P(2,8) is horizontal.