Rotating an Ellipse to its original orientation

Question

I have this equation. A rotated ellipse.

$37x^{2}+42\sqrt{3}xy+79y^{2}-400=0$

Easy to see that

$\tan\left(2\theta\right)=-\sqrt{3}$

and when I rotate it counter clockwise 30 degrees, I get,

$4x^{2}+25y^{2}=100$

which is the wrong answer. 3 versions of the book we use for precalculus fail to mention that if the cotangent (and therefore tangent) is negative, one should treat the angle as it were occuring in the second quadrant. Live and learn, and get a good result,

$25x^{2}+4y^{2}=100$

Now, as I prepare to be in school and answer questions about this process, I am at a loss as to how to create a rotation whose equation leads back to an original "X is major axis" ellipse. To be clear, when I rotate

$4x^{2}+25y^{2}=100$

Clockwise 30 degrees or counterclockwise 150, the equation of the tilted ellipse leads back to the y-axis major axis orientation. Clearly, I am missing something.

To add - we are given the Red ellipse equation, and are now able to unrotate to the Blue. I would simply like to know what it would take to produce an ellipse that unrotates to the Green orientation. I will edit in any details that need clarifying.

Edit - I understand and calculated the choice of 60, 150 as rotations. The worksheet gave only 60 as the answer. An online math teacher did a very similar problem and posted this warning:

Which led to my question.

Answer 1

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.

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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 $0^\circ\lt \theta\lt 180^\circ$.

Answer 2

Well, clearly rotating an ellipse about its center counterclockwise 30° or clockwise 150° must have the same result, due to the symmetry of the curve. Those rotations are separated by 180°.

You say that if the tangent of an angle is negative, the angle must be in quadrant II. Why can it not be in quadrant IV? I guess you are referring to the angle $2\theta$. Consider this:

$\tan(2\theta)=-\sqrt{3}$

$2\theta=120°$

$\theta=60°$

or this:

$\tan(2\theta)=-\sqrt{3}$

$2\theta=300°$

$\theta=150°$

I would not call either answer wrong unless it conflicts with the instructions of the problem.