# Thread: points on a curve

1. ## points on a curve

given y^2= 2+xy and dy/dx= y/2y-x find all points (x,y) on the curve where the line tangent to the curve has slope 1/2

Do you need both equations to solve? I tried solving it and got x=0 and
y=sq.rt. of 2

2. Originally Posted by wisezeta
given y^2= 2+xy and dy/dx= y/2y-x find all points (x,y) on the curve where the line tangent to the curve has slope 1/2

Do you need both equations to solve? <<<<< Yes
I tried solving it and got x=0 and $\displaystyle y=\sqrt{2}$ <<<<< Correct
Don't forget that there exist a second result for y! That means you'll get 2 parallel tangents.

3. Ok so if i was proving that none were present on the horizontal line i would set it equal to 0. Would the proof be y=0 for the derivative equation and 0=2 for the original, which proves it since the two numbers are not equal?

4. Originally Posted by wisezeta
Ok so if i was proving that none were present on the horizontal line i would set it equal to 0. Would the proof be y=0 for the derivative equation and 0=2 for the original, which proves it since the two numbers are not equal?
In my opinion that must be sufficient.

To use another attempt you can prove that the graph has a horizontal asymptote y = 0.

5. I was thinking something similar. Thank you very much.