calculus
posted by anonymous on .
x^2x^2y=y^21
find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

sub x = 2 into your equation to get
4  4y = y^2  1
0 = y^2 + 4y  5
(y+5)(y1) = 0
so when x=2, y =5 or when x=2, y = 1
so the points are (2,5) and (2,1)
check them by subbing them back into the original equation, they work 
I found the derivative of your equation implicitly and got
dy/dx = (2x+2xy)/(2y = x^2)
for a horizontal tangent this has to be zero, which means the numerator should be zero
subbing in (2,5) makes the top zero, so there is a horizontal tangent at the point (2,_5)