Question 455464: find the points on the curve x^2-xy+y^2=48 at which the tangents are parallel and perpendicular to x- axis Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! find the points on the curve x^2-xy+y^2=48 at which the tangents are parallel and perpendicular to x- axis
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Parallel is slope = 0
Perpendicular --> dx/dy = 0
2x*dx - y*dx - x*dy + 2y*dy = 0
(2x - y)*dx + (-x + 2y)*dy = 0
(2x - y)*dx = (x - 2y)*dy
dy/dx = (2x-y)/(x-2y)
Parallel:
dy/dx = (2x-y)/(x-2y) = 0
y = 2x
Sub into x^2-xy+y^2=48
x = 4, -4
--> (-4,-8) and (4,8)
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Perpendicular:
dx/dy = (x-2y)/(2x-y) = 0
--> (-8,-4) and (8,4)
