The temp. in (x, y) is 4*x^2+24*y^2-2*x.
From all the ellipse points 4*x^2+12*y^2 = 1 find the hotest points and coldest points.
I am just rewriting for my own clarification.
$T(x,y) = 4x^2+24y^2-2x$
$4x^2+12y^2=1$ This tells us that $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$ (I'm not going to worry about the range for y for now. We can revisit that later).
So, we want to simplify the function a bit. Let's solve for $y^2$ to get a function of one variable.
$4x^2+12y^2 = 1$
$12y^2 = 1-4x^2$
$y^2 = \dfrac{1-4x^2}{12}$
Plugging into the formula for $T$, we get:
$T(x) = 4x^2+24\left(\dfrac{1-4x^2}{12}\right)-2x$
$T(x) = 4x^2+2-8x^2-2x$
$T(x) = 2-2x-4x^2 = -4\left(x^2+\dfrac{1}{2}x- \dfrac{1}{2}\right)$
Now, we can complete the square.
$T(x) = -4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)-3$
$T(x) = -4\left(x+\dfrac{1}{2}\right)^2-3$
This is a parabola. It reaches its maximum when $x = -\dfrac{1}{2}$. There are at most two values for $y$ for this value of $x$.
It reaches its minimum when $x$ is as far from $-\dfrac{1}{2}$ as possible. The only value where that can occur is $\dfrac{1}{2}$. There are at most two values for $y$ for this value of $x$.
Can you find the two or four points, one or two where the temperature is hottest and one or two where it is coldest?