find the equation 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 = 0
I don't know what to do after this
4 answers
you know that y^2 = (x^2+4)/x, so just plug that in and solve for x.
wouldn't everthing cancel?
2x - (x^2+4)/x = 0
2x^2 - x^2 - 4 = 0
x^2 - 4 = 0
x = -2 or 2
Now, find y at those values, and those are the horizontal tangents.
2x^2 - x^2 - 4 = 0
x^2 - 4 = 0
x = -2 or 2
Now, find y at those values, and those are the horizontal tangents.
2x - ( x^2 + 4)/x = 0
times x
2x^2 -(x^2 + 4) = 0
x^2 = 4
x = ± 2
if x = 2, y^2 = 4, and dy/dx = 0 , of course !
a horizontal line through (2,4) is y = 4
if x = -2 ....... (you do it)
verification of result at
http://www.wolframalpha.com/input/?i=plot+y%5E2+%3D+%28x%5E2%2B4%29%2Fx
times x
2x^2 -(x^2 + 4) = 0
x^2 = 4
x = ± 2
if x = 2, y^2 = 4, and dy/dx = 0 , of course !
a horizontal line through (2,4) is y = 4
if x = -2 ....... (you do it)
verification of result at
http://www.wolframalpha.com/input/?i=plot+y%5E2+%3D+%28x%5E2%2B4%29%2Fx
1 answer