waht is the slope of the normal line to y^2 = x/2 at P(1/8,1/4)??
Janice, Nadine, Elmo, Keisha and Amber, you're probably working together one way or another. So you can also help each other.
As before, use implicit differentiation to get y':
2yy'=(1/2)
y'=(1/2)/(2y)
The slope of the normal is given by
m=-1/y'=-4y
So the line for the normal is
L1: (y-y1)=m(x-x1)