Sketching for isotherms $T(x, y) = \text{ const}$ Hi I came across this question 

Let the temperature $$T(x,y) = 4x^2+16y^2$$ in a body be independent of $z$, identify the isotherms $T(x, y) = \text{ const}$. Sketch it.

I do not understand the question. What type of equation(s) do I have to sketch?
 A: You have to sketch
$$4x^2+16y^2=c, \text{ where $c>0$ is a constant number.}$$
that is equivalent to
$$\frac{x^2}{(\sqrt{c}/2)^2}+\frac{y^2}{(\sqrt{c}/4)^2}=1$$
which is an ellipse.
If $c=0$ you just have one point which is $(x,y)=(0,0)$.
If $c<0$ we don't have any isotherm.
A: For several different values of $C$, you should draw lines containing all the points $(x, y)$ that solve $4x^2 + 16y^2 = C$, and you should also describe what the isotherms look like for a general value of $C$. For example, if $C = 16$, then the graph you would draw is $\frac{x^2}{4} + y^2 = 1$ (this is an ellipse with center at the origin, semi-major axis $2$, and semi-minor axis $1$).
A: if you impose $$T\left(x,y\right) = C = \mathrm{cte}$$ you will get a set of points $\left(x,y\right)$ that satisfy it (a curve in the $Oxy$ plane).
For each different $C$ you will get a different curve. In that curve the temperature will be constant and equal to $C$. 
In this case, your equation for the curves would be:
$$ 4y^2 + 16y^2 = C$$
as $C$ is a temperature it has to be $C\ge0$ and you can rewrite your equation like
$$ \left(\frac{x}{\frac{\sqrt{C}}{2}}\right)^2+\left(\frac{y}{\frac{\sqrt{C}}{4}}\right)^2= 1$$
and I invite you to look for the equation of an ellipse on the wikipedia.
