# Find the coordinates of all points

1. Dec 12, 2005

### force

without using a graphing calculator
---------------------------------
How would you find the coordinates of all points(x,y) of the tangent lines of a relation such as (x^2)*y - y^3 = 8 (btw this graph looks like 3 parabolas oppening, one to the right, left and one downwards) where the tangent line is horzontal ?

Would using implicit differentiation work getting y'=(-2xy)/((x^2) -3y^2) then equating it to zero(slope) ?

Consider x^3 + y^3 - 9xy = 0 how can you isolate the y on one side in the form y = x ?

2. Dec 13, 2005

### benorin

Consider the curve given implicitly by:

$$x^{2}y-y^{3}=8$$

and the derivative

$$y^{\prime}=\frac{-2xy}{x^{2}-3y^{2}}$$

horizontal tangent lines occur when $y^{\prime}=0$, so set $-2xy^2=0 \Rightarrow x=0\mbox{ or }y=0$, and we know that $y\neq 0$ at all points on the curve, but $x=0$ is a point on the curve, namely (0,-2), where we do indeed have a horizontal tangent line to the curve.