Question

x^2-x^2y=y^2-1

find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

Answer 1

sub x = 2 into your equation to get

4 - 4y = y^2 - 1
0 = y^2 + 4y - 5
(y+5)(y-1) = 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

Answer 2

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)