Question related to ellipses and their properties

Question

I was solving this question related to ellipses but got stuck, please help me in angle chasing:

Could not even move further :

Consider the ellipse $4x^2+y^2-8x+4y+4=0.$ Pair of tangents $PT_1$ and $PT_2$ are drawn to the ellipse from $P(-8,-2).$ Let $S_1'$ be the image of $S_1$ about $PT_1$ and $S_2'$ be the image of $S_2$ about $PT_2$, then if $\angle PS_1S_2'=\alpha$ and $\angle PS_1'S_2=\beta$, then $\displaystyle\frac{\alpha}{\beta}$ is equal to $\ldots$? ($S_1, S_2$ refer to the foci of the ellipse.)

My progress in an image:

I only want mathematical and rigorous proofs as it is also very much self-evident to me that and equal by symmetry .

Answer 1

For those looking for the easier answer :

Told to me by someone :

$PS_1 =PS_2$

$PS_1 ^{'} = PS_2 ^{'}$

$PS_1 = PS_1 ^{'}$(Congruent triangles) , $PS_2 = PS_2^{'}$

Now $S_1S_2^{'} = 2a$, as $S_1S_2^{'} = S_2^{'}T_2 + S_1T_2$ (Property that $S_2^{'} ,T_2$ and $S_1$ are collinear)

$S_2^{'}T_2 = S_2T_2$

$S_1S_2^{'} = S_2T_2 + S_2T_2 =2a$

$S_2S_1^{'} = 2a $

Now we can say that by congruent triangles

Both angles are equal