Graphing Implicit Function: Find HTL & VTL

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Graphing Implicit Function

Homework Statement

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.

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.

 

[LCKurtz]

Homework Statement

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.

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.

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.

 

[Light bulb]

alright ill give it a try

 

[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 won't give me an answers, I've looked at the graph and there are HTL at (+-.66,+-1.16) and VTL at (+-1.16,+-.66)

 

[HallsofIvy]

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.

 

[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.

 

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ahh ill give it another try, thanks guys

 

[HallsofIvy]

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

 

[Light bulb]

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