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!