Calculus
posted by W on .
If 3x^22y^2=4xy + 20, find the point(s) at which the graph from Problem 28 has vertical tangents. I found that dy/dx= (3x2y)/(2x+2y), but now I don't know what to do. Do I set dy/dx to 0?

For a vertical line, the slope would be undefined, that is,
(3x2y)/(2x+2y) has to be undefined.
When does that happen?
When the denominator is zero, so
2x + 2y = 0
y = x
sub into the origianal
3x^2  2(x)^2 = 4x(x) + 20
5x^2 = 20
x^2=4
x = ±2
if x=2, y = 2,
if x = 2, y = 2
points where tangent is vertical are (2,2) and (2,2) 
thanks! that was really helpful ;)