At what point(s) does the curve defined by x^2-6x=6-6y^2 have horizontal tangents?
I took the derivative and solved for y'. I got y'=-1/6x/y+1/2y. hmm...what am I suppose to do?
leave your derivative as
dy/dx = (2x-6)/(-12y) after you differentiated implicitly.
then (2x-6)/(-12y) = 0 can only be true if
2x-6 = 0
x = 3 now put that back into the original equation to get y = ...
your equation will be y = that value
would it by (2,sqrt(10)/2)?
To find the points on the curve where the tangent line is horizontal, we need to set the derivative, y', equal to zero.
You correctly found the derivative as y' = (-1/6x/y) + (1/2y). To find when y' = 0, we set the derivative equal to zero:
(-1/6x/y) + (1/2y) = 0
Now, let's simplify the equation to solve for x and y. Multiply through by 6xy to eliminate the denominators:
-xy + 3x^2y = 0
Rearrange the terms:
3x^2y = xy
Now, we can divide both sides by xy, assuming xy is not zero:
3x^2 = x
Next, divide through by x to isolate x:
3x = 1
Finally, solve for x:
x = 1/3
To find the corresponding y-values, substitute the value of x back into the original equation:
x^2 - 6x = 6 - 6y^2
(1/3)^2 - 6(1/3) = 6 - 6y^2
1/9 - 2 = 6 - 6y^2
-17/9 = -6y^2
Now solve for y:
y^2 = 17/54
y = ±√(17/54)
Thus, the curve defined by x^2 - 6x = 6 - 6y^2 has horizontal tangents at the points (1/3, √(17/54)) and (1/3, -√(17/54)).