# Implicit Graphing

1. Nov 6, 2009

### Light bulb

Graphing Implicit Function

1. The problem statement, all variables and given/known data

Graph (x^2+Y^2)^2=4xy
A)Find Y'
B)Find all points that have vertical tangent lines.
C)Find all points that have horizontal tangent lines.

3. The attempt at a solution

I feel like I'm going nuts on this problem, y' is (-x^3-xy^2+y)/(y^3+x^2y-x) right?
I used a graphing program and it looks like an 8 reflected over y=x, but I just can't understand why I'm not getting the right answers for the HTL and the VTL.

Last edited: Nov 6, 2009
2. Nov 6, 2009

### LCKurtz

Your derivative is correct. Try factoring an x out of the first two terms of the numerator and a y out of the first two terms of the denominator and replace the x2 + y2 with square root of 4xy from the original equation and see if that helps.

3. Nov 7, 2009

### Light bulb

alright ill give it a try

4. Nov 7, 2009

### Light bulb

Im still getting stumped, if I do that, I can turn the top half into y=4x^3, and the bottom half into x=4y^3, but obviously those wont give me an answers, Ive looked at the graph and there are HTL at (+-.66,+-1.16) and VTL at (+-1.16,+-.66)

5. Nov 7, 2009

### HallsofIvy

Staff Emeritus
Vertical tangent lines are where the denominator of the derivative, $y^3+x^2y-x$ is 0 and horizontal tangent lines are where the numerator of the derivative, $-x^3-xy^2+y$ is 0. For each of those, you still have $(x^2+y^2)^2=4xy$ so in each case you have two equations to solve for x and y.

For vertical tangent lines solve the system
$y^3+x^2y-x= 0$
$(x^2+y^2)^2=4xy$.

For horizontal tangent lines solve the system
$-x^3-xy^2+y= 0$
$(x^2+y^2)^2=4xy$.

Last edited: Nov 7, 2009
6. Nov 7, 2009

### Mentallic

You're on the right track. For the horizontal asymptote given as $y=4x^3$ you need to solve is simultaneously with the curve $(x^2+y^2)^2=4xy$ as hallsofivy has said.
Same thing goes for the vertical asymptote.

7. Nov 7, 2009

### Light bulb

ahh ill give it another try, thanks guys

8. Nov 7, 2009

### HallsofIvy

Staff Emeritus
A little additional point: The first equation for vertical tangent lines can be written as $y^3+ x^2y- x= y(y^2+ x^2)- y= 0$ so $x^2+ y^2= y/y= 1$

9. Nov 7, 2009

### Light bulb

I had to talk a little break from it but I just came back to it and got it right, thanks pf!