Sketching level curves of f(x,y)

[username12345]

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.

[CompuChip]

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) ?

[username12345]

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

???

[CompuChip]

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.

[username12345]

Ok, that explanation helped, thankyou.