# Find dy/dx and then find slope of tangent(s).

1. Jan 25, 2012

### Trevor404

1. The problem statement, all variables and given/known data
a) If 2y2-xy-x2=0 , find dy/dx.
b) What is/are the slope(s) of the tangent(s)at x=1?

2. Relevant equations

3. The attempt at a solution
Using implicit differentiation:

a) (4y)dy/dx-(x)dy/dx-y(1)-2x=0

(4y)dy/dx-(x)dy/dx=2x+y

(4y-x)dy/dx=2x+y

dy/dx=(2x+y)/(4y-x)

Assuming that I found dy/dx correctly (please let me know if I did not) I am having some trouble seeing where to go from here to complete part b). If I had x and y coordinates I could work through it but with only an x value it is proving difficult for me.

My latest attempt was to try and solve for y using the original equation.

b) 2y2-xy-x2=0

At x=1, 2y2-(1)y-(1)2=0

Leaving me with:
2y2-y=1

And that's where I ended up, stuck and admittedly confused.
Any help or nudges in the right direction would be greatly appreciated.

Thanks,
-Trevor

2. Jan 25, 2012

### vela

Staff Emeritus
Hint: You have a quadratic equation, 2y2-y-1=0.

3. Jan 25, 2012

### Trevor404

Ah yes I do!
So from 2y2-y-1=0
I get (2y+1)(y-1)=0

I can use this to find my y coordinates and then use the x and y values with my previously found dy/dx to get my slope at both points, correct?
I calculated a slope of 1 at (1,1) and a slope of -1/2 at (1,-1/2).

Thanks a lot!