Let $P := (a \cos\phi, b \sin\phi)$ on an origin-centered ellipse with radii $a$ and $b$; define $c := \sqrt{a^2-b^2}$, so that the ellipse's eccentricity is $e := c/a$. The line through $P$, normal to the ellipse —that is, in direction $(b\cos\phi,a\sin\phi)$— meets the $x$-axis at $K:= (k,0)$, where $k:= c^2/a \cos\phi$. So, $K$ is the center of a circle internally tangent to the ellipse at $P$, and its radius, $r$, is given by
$$r^2 = |PK|^2 = \frac{b^2(a^2-c^2\cos^2\phi)}{a^2} = \frac{b^2(c^2-k^2)}{c^2} \tag{1}$$
so that
$$\frac{r^2}{b^2}+\frac{k^2}{c^2}=1 \tag{2}$$
This allow us to write, for some $\theta$,
$$r = b\sin\theta \qquad k = c \cos\theta \tag{3}$$
Now, suppose $\bigcirc K_0$ and $\bigcirc K_1$ are circles internally tangent to the ellipse, with respective centers and radii given by $(3)$ for $\theta = \theta_0$ and $\theta=\theta_1$. If these circles are tangent to each other (with $K_1$ "on the right" of $K_0$), then
$$\begin{align}
k_0 + r_0 &= k_1 - r_1 \\[4pt]
\to\quad -2 c \sin\frac{\theta_0 + \theta_1}{2} \sin\frac{\theta_0 - \theta_1}{2} &= -2 b \sin\frac{\theta_0 + \theta_1}{2} \cos\frac{\theta_0 - \theta_1}{2} \\[6pt]
\to\quad \tan\frac{\theta_0 - \theta_1}{2} &= \frac{b}{c} \\[6pt]
\to\quad \theta_1 &= \theta_0 - 2\arctan\frac{b}{c} \\[6pt]
&= \theta_0 - 2\arccos e \tag{4}
\end{align}$$
More generally, if circles $\bigcirc K_i$, defined by $\theta = \theta_i$ in $(3)$, form a tangent chain, then
$$\theta_i = \theta_0 - 2 i \arccos e \tag{5}$$
where index $i$ is subject to certain viability conditions (eg, $\theta_i \geq 0$) that we'll assume hold. Thus, defining $\varepsilon := 2\arccos e$, we have
$$\begin{align}
\frac{r_{i+j} + r_{i-j}}{r_i} &= \frac{b\sin(\theta_0-(i+j)\psi)+b\sin(\theta_0-(i-j)\varepsilon)}{b \sin(\theta_0-i\varepsilon)} \\[6pt]
&= 2\cos j\varepsilon = 2\cos( 2j \arccos e ) \\[4pt]
&= 2\,T_{2j}(e) \tag{6}
\end{align}$$
where $T_{2j}$ is the $2j$-th Chebyshev polynomial of the first kind. Notably, the value of $(6)$ is independent of $i$. In particular, if we take $j=3$ and both $i=4$ and $i=7$, we can write
$$\frac{r_{4-3}+r_{4+3}}{r_4} = 2\;T_{2\cdot 3}(e) =\frac{r_{7-3}+r_{7+3}}{r_7} \tag{7}$$
which gives the result. $\square$
Addendum. In this follow-up question, @g.kov asks when an ellipse allows a "perfect packing" of $n$ tangent circles along its axis. It seems reasonable to append here a justification of the condition given there.
In a perfect packing, the first and last circles in a chain are tangent to the ellipse at the endpoints of the axis, so that their radii match the ellipse's radius of curvature (namely, $b^2/a$) at those points. Thus, we have
$$r_0 = r_{n-1} = \frac{b^2}{a} \quad\to\quad \sin\theta_0 = \sin\theta_{n-1} = \frac{b}{a} \quad\to\quad \cos\theta_0 = \cos\theta_{n-1} = e \tag{8}$$
We can say that $\theta_0 = \pi - \arccos e$ and $\theta_{n-1} = \arccos e$. By $(5)$, this implies
$$\arccos e = \theta_{n-1} = \theta_0 - 2(n-1)\arccos e = (\pi - \arccos e) - 2(n-1)\arccos e \tag{9}$$
so that
$$\pi = 2n\arccos e \qquad\to\qquad \cos \frac{\pi}{2n} = e \tag{10}$$
This is equivalent to @g.kov's condition for a perfectly-packable ellipse. $\square$
Best Answer
HINITS
Part a) of the problem is to put the equation $$13y^2 + 7x^2+6\sqrt{3}xy+4y-4\sqrt{3}x=0$$ in so-called standard form, by rotating the axes and translating the origin. It's hard to know what advice to give you on this, because you haven't indicated how much of this topic you understand.
When you look on the internet, you see lots of examples where the original equation doesn't have an $xy-$term. In that case, rotation of the axes is not necessary. Do you know how to deal with these problems when the $xy-$ term is lacking? (The method is called "completing the square.")
The first step in doing these problems is to eliminate the $xy-$ term, and then to eliminate the $x-$ and $y-$ terms leaving an equation of the form $$\left({x-a\over A}\right)^2+\left({y-b\over B}\right)^2=1$$
If you haven't practiced eliminating the linear terms, that is the $x-$ and $y-$ terms from equations lacking an $xy-$term yet, I suggest you put this problem aside for the moment, and try the easier problems first.
EDIT
Here is a brief introduction to rotation of axes, but it really says everything you need to know. If that's not enough, Google "rotation of axes in an ellipse" and you'll get lots of hits. The best way to use this site it to show us how far you've gotten, and where you'e stuck. Then we'll be able to give you advice that's appropriate for you individually.