# Sketching level curves of f(x,y)

48
1. The problem statement, all variables and given/known data

Sketch the level curve of the surface $$z = \frac{x^2 - 2y + 6}{3x^2 + y}$$ belonging to height z = 1 indicating the points at which the curves cut the y−axis.

2. Relevant equations

3. The attempt at a solution

I put $$1 = \frac{x^2 - 2y + 6}{3x^2 + y}$$ but then don't know how to proceed.

The answer shows an inverted parabola at y = 2, but I don't know how to get that.

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3. ### CompuChip

4,299
From
$$1 = \frac{x^2 - 2y + 6}{3x^2 + y}$$
it follows that
$$3x^2 + y = x^2 - 2y + 6$$

Can you cast this in the form
y = f(x) ?

48
Do you mean like this...

$$3x^2 + y = x^2 - 2y + 6$$
$$3y = -2x^2 + 6$$
$$y = \frac{-2x^2 + 6}{3}$$
$$y = \frac{-2x^2}{3} + 2$$

???

5. ### CompuChip

4,299
Yes.
Now, that's a parabola, isn't it? It's of the general form y = ax^2 + bx + c with a = -2/3, b = 0, c = 2; you can see that it is inverted (like a mountain top) because a < 0.