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Homework Statement
ƒ(x,y) = ln(x^{2}+4y^{2})
Homework Equations
I'm not really sure but I solved for y
The Attempt at a Solution
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Yes, that was a bit confusing in the OP, but judging from the drawing I would say cosmos42 understood about different values for C."Level curves" are curves on which the expression has a specific value. I would not solve for y. Instead, I would look at the equation [itex]ln(x^2+ 4y^2)= C[/itex] and recognize that, taking the exponential of both sides, [itex]x^2+ 4y^2= e^C[/itex] which is equivalent to [itex]\frac{x^2}{e^C}+ \frac{y^2}{e^C/4}= 1[/itex]. You should immediately see that, for every C, this is an ellipse, centered at the origin, with x intercepts at [itex]\left(-e^{C/2}, 0\right)[/itex] and [itex]\left(e^{C/2}, 0\right)[/itex], y intercepts at [itex]\left(0, -\frac{e^{C/2}}{2}\right)[/itex] and [itex]\left(0, \frac{e^{C/2}}{2}\right)[/itex].