Posted by **W** on Wednesday, December 14, 2011 at 9:26pm.

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

- Calculus -
**Reiny**, Wednesday, December 14, 2011 at 11:29pm
For a vertical line, the slope would be undefined, that is,

(3x-2y)/(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)

- Calculus -
**W**, Wednesday, December 14, 2011 at 11:32pm
thanks! that was really helpful ;)

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