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First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2a\cos\theta$ (where $a$ is its radius).

If you have an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=a\cos\theta$, $y=b\sin\theta$, and you get a modified polar coordinate system by "filling it in" with $x=ar\cos\theta$, $y=ar\sin\theta$. (Note that $\theta$ is no longer the angle between the $x$-axis and the ray to the point.)

Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation $$(x+y)^4 = 4xy.$$ I'm not sure I get the point of introducing $\alpha$. The equation of this curve in usual polar coordinates would be $$r^2 = \frac{2\sin 2\theta}{(\cos\theta+\sin\theta)^4},$$ with $0\le\theta\le\pi/2$ or $\pi\le\theta\le 3\pi/2$ (to make $\sin 2\theta\ge 0$). If we use $\alpha = 2$, with $x=r\cos 2\theta$ and $y=r\sin 2\theta$, this equation becomes $$r^2 = \frac{2\sin 4\theta}{(\cos 2\theta + \sin 2\theta)^4},$$ with $0\le\theta\le \pi/4$ or $\pi/2\le\theta\le 3\pi/4$. Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $\alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)

So, how would we try to simplify either expression? We note that $$\cos\theta+\sin\theta = \sqrt2\sin\big(\theta+\frac{\pi}4\big),$$ so we want to substitute $\psi = \theta+\frac{\pi}4$. Then and then $\sin 2\theta = \sin 2(\psi-\frac{\pi}4)=\sin(2\psi - \frac{\pi}2) = -\cos 2\psi$. Then we end up with $$r^2 = \frac{-\cos 2\psi}{2\sin^4\psi}, \quad \pi/4\le\psi\le 3\pi/4 \quad\text{or}\quad 5\pi/4\le\psi\le 7\pi/4.$$ Now let's try this with the second formulation. Analogously, we have $$\cos 2\theta+\sin 2\theta = \sqrt2\sin\big(2\theta+\frac{\pi}4\big) = \sqrt2\sin 2\big(\theta+\frac{\pi}8\big).$$ This seems very unhelpful. Perhaps we should try scaling with $\alpha = 1/2$? Then we'd have $$\sin\frac{\theta}2 + \cos\frac{\theta}2 = \sqrt2\sin\big(\frac{\theta}2 +\frac{\pi}4\big) = \sqrt2\sin\frac12\big(\theta+\frac{\pi}2\big).$$ Does that seem a bit more promising?

At any rate, I am not sure I believe your book is right. A translation of $\theta$ is clearly called for before a rescaling of $\theta$.