I think that the warning means that if $\cot(2\theta)\gt 0$, then you can find the smallest positive $\theta$ in $0^\circ\lt\theta\lt 45^\circ$ such that the ellipse you get has no $xy$ term, and if $\cot(2\theta)\lt 0$, then you can find the smallest positive $\theta$ in $45^\circ\lt\theta\lt 90^\circ$ such that the ellipse you get has no $xy$ term. I think that it does not say anything about whether the ellipse you get is "X is major axis" ellipse or not. I think that it means that for such a $\theta$, you get an ellipse which has no $xy$ term, i.e. whose axis is either X-axis or Y-axis.
In the following, I'll show the condition that the ellipse you get is "X is major axis" ellipse.
If one rotates an ellipse $ax^2+2bxy+cy^2+g=0$ where $$a\gt 0,c\gt 0,c\not=a,ac-b^2\gt 0,g\lt 0$$ about the origin counter clockwise $(\color{red}−θ)$ where $0^\circ\lt\theta\lt 180^\circ$, then one has
$$a(x\cos(-\theta)+y\sin(-\theta))^2+2b\sqrt 3(x\cos(-\theta)+y\sin(-\theta))(-x\sin(-\theta)+y\cos(-\theta))+c(-x\sin(-\theta)+y\cos(-\theta))^2+g=0,$$
i.e.
$$(a\cos^2(-\theta)-2b\cos(-\theta)\sin(-\theta)+c\sin^2(-\theta))x^2$$$$+(2a\cos(-\theta)\sin(-\theta)+2b\cos^2(-\theta)-2b\sin^2(-\theta)-2c\sin(-\theta)\cos(-\theta))xy$$$$+(a\sin^2(-\theta)+2b\sin(-\theta)\cos(-\theta)+c\cos^2(-\theta))y^2+g=0,$$
i.e.
$$(b\sin(2\theta)+\frac{a-c}{2}\cos(2\theta)+\frac{a+c}{2})x^2+((c-a)\sin(2\theta)+2b\cos(2\theta))xy+(\frac{c-a}{2}\cos(2\theta)-b\sin(2\theta)+\frac{a+c}{2})y^2+g=0\tag1$$
Now, $$(c-a)\sin(2\theta)+2b\cos(2\theta)=0$$
is equivalent to
$$\tan(2\theta)=\frac{-2b}{c-a}\tag2$$
Under $(2)$, $(1)$ becomes
$$\frac{x^2}{\dfrac{-g}{b\sin(2\theta)+\frac{a-c}{2}\cos(2\theta)+\frac{a+c}{2}}}+\frac{y^2}{\dfrac{-g}{\frac{c-a}{2}\cos(2\theta)-b\sin(2\theta)+\frac{a+c}{2}}}=1$$
So, the condition that one gets "X is major axis" ellipse is
$$(2)\quad\text{and}\quad \dfrac{-g}{b\sin(2\theta)+\frac{a-c}{2}\cos(2\theta)+\frac{a+c}{2}}\gt \dfrac{-g}{\frac{c-a}{2}\cos(2\theta)-b\sin(2\theta)+\frac{a+c}{2}}$$
i.e.
$$(2)\quad\text{and}\quad \frac{c-a}{2}\cos(2\theta)-b\sin(2\theta)+\frac{a+c}{2}\gt b\sin(2\theta)+\frac{a-c}{2}\cos(2\theta)+\frac{a+c}{2}$$
i.e.
$$(2)\quad\text{and}\quad (c-a)\cos(2\theta)\gt 2b\sin(2\theta)$$
i.e.
$$(2)\quad\text{and}\quad (c-a)\cos(2\theta)\gt 2b\frac{-2b}{c-a}\cos(2\theta)$$
i.e.
$$(2)\quad\text{and}\quad \frac{(c-a)^2+4b^2}{c-a}\cos(2\theta)\gt 0$$
i.e.
$$(2)\quad\text{and}\quad \frac{\cos(2\theta)}{c-a}\gt 0$$
So, the condition that one gets "X is major axis" ellipse is
$$\begin{cases}(2)\quad\text{and}\quad \cos(2\theta)\gt 0&\text{if $c\gt a$}
\\\\(2)\quad\text{and}\quad \cos(2\theta)\lt 0&\text{if $c\lt a$}
\end{cases}$$
In our case where $$a=37,2b=42\sqrt 3,c=79,g=-400$$
the condition that one gets "X is major axis" ellipse is
$$\tan(2\theta)=-\sqrt 3\quad\text{and}\quad \cos(2\theta)\gt 0$$
i.e.
$$\theta=150^\circ$$
under the condition that $0^\circ\lt \theta\lt 180^\circ$.
Best Answer
It is a minor detail: Let $x$ be the price, as above. Then it is known that $\$ 25$ is taken off, and the price is now $150$. So it must be that $x - 25 = 150$, and consequently, $x = 175$.
The distinction is that your son has jumped a step in the reasoning. You do know at the very beginning of the problem that $x = 25 + 150$, you only know that $x - 25 = 150$.
I hope this helps.