Question
find the equations of ALL horizontal tangents to the curve y^2=(x^2+4)/x, if any exist
This is what I have so far:
xy^2= x^2+4
2xyDy/Dx =2x-y^2
dy/dx = (2x- y^2)/2xy
(2x- y^2)/2xy = 0
2x-y^2
I don't know what to do next
This is what I have so far:
xy^2= x^2+4
2xyDy/Dx =2x-y^2
dy/dx = (2x- y^2)/2xy
(2x- y^2)/2xy = 0
2x-y^2
I don't know what to do next
Answers
Steve
So far, so good. We know that for a horizontal tangent, y'=0, so
2x-y^2 = 0
But, we already know that
y^2 = (x^2+4)/x, so
2x - (x^2+4)/x = 0
2x^2 - x^2-4 = 0
x^2 = 4
x = 2 or -2
At x = 2, y = 4
At x = -2, y = -4
Those are the horizontal tangents
2x-y^2 = 0
But, we already know that
y^2 = (x^2+4)/x, so
2x - (x^2+4)/x = 0
2x^2 - x^2-4 = 0
x^2 = 4
x = 2 or -2
At x = 2, y = 4
At x = -2, y = -4
Those are the horizontal tangents