## Need help in drawing a level curve.

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

i want to draw a level curve for the following but im having some trouble.

x-y^2/x^2+y

2. Relevant equations

3. The attempt at a solution

I know i'm supposed to set c = x-y^2/x^2+y, but i can't find a way to set y in terms of x and c.

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Mentor
 Quote by AndreTheGiant 1. The problem statement, all variables and given/known data i want to draw a level curve for the following but im having some trouble. x-y^2/x^2+y 2. Relevant equations 3. The attempt at a solution I know i'm supposed to set c = x-y^2/x^2+y, but i can't find a way to set y in terms of x and c.
Let me see if I have this straight. Your function is f(x, y) = x - (y2/x2) + y, right? That's what you wrote.

 Whoops, guess I should've added brackets. [x-y^2]/[x^2+y] sorry bout that.

Mentor

## Need help in drawing a level curve.

Much better.

Your equation is z = (x - y2)/(x2 + y)

The level curves are curves in some plane, z = c.

For example, in the plane z = 1, the level curve is the equation (x - y2)/(x2 + y) = 1. Keep in mind that y can't equal -x2.

Multiply both sides by x2 + y. What equation do you get?

 c(x^2+y) = (x-y^2)?
 Mentor So if you move all of the terms to one side, what do you get?
 cx^2 + cy - x + y^2 = 0 so when c = 0 i get a parabola, but what about when c = 2 for example..?

Mentor
 Quote by AndreTheGiant cx^2 + cy - x + y^2 = 0 so when c = 0 i get a parabola, but what about when c = 2 for example..?
Complete the squares.

2x2 - x + y2 +2y = 0

2(x2 - (1/2)x) + y2 +2y = 0

2(x2 - 2 (1/4)x + (1/4)2 -1/16 ) + y2 +2y + 1 - 1 = 0

2(x - 1/4)2 + (y - 1)2 = 1 + 1/8

(x - 1/4)2/(9/16) + (y - 1)2/(9/8) = 1

Do similar for c in general.

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