1. Given hyperbola and its foci

The hyperbola is

So

Hence the foci are

Let

which is the standard parametrization for the first quadrant point on the hyperbola.

2. Tangent slope at $P$

Differentiate $\frac{x^2}{100}-\frac{y^2}{64}=1$ to get

so

At $P=(10\sec\theta,8\tan\theta)$,

Thus the required line through $S=(c,0)$ is $y=m(x-c).$ Call this line $SP_1$.

3. Distance $\delta$ from $P$ to line $SP_1$

Equation of line $SP_1$: $y-mx+mc=0.$ Distance from $P=(x_P,y_P)$ to this line is

Now

Compute:

But

so

Hence

Therefore

Since $c=2\sqrt{41}$,

So

Multiply numerator and denominator suitably:

Thus

But

Hence

Factor the denominator:

So

4. Compute $\beta=S_1P$

Using distance formula from $S_1=(-c,0)$ to $P=(10\sec\theta,8\tan\theta)$,

Expand:

Using $c^2=164$ and $\tan^2\theta=\sec^2\theta-1$,

Since $c=2\sqrt{41}$,

so

Therefore

A more useful factorized form is

5. Compute $\sin\dfrac\alpha2$

For a hyperbola, the difference of focal distances is constant:

Thus if

then

In triangle $SPS_1$, side $SS_1=2c=4\sqrt{41}$ and angle at $P$ is $\alpha$.

Use the half-angle identity:

where the sides adjacent to angle $\alpha$ are

and opposite side is

Semiperimeter:

Then

Therefore

So

Now

Hence

Thus

Again using factorization,

6. Now compute the target expression

We need

First compute $\beta\delta$:

So

Multiply by $\sin\dfrac\alpha2$:

Therefore,

Hence the greatest integer less than or equal to this is

7. Comparison with stored answer

Stored correct answer: $7$

Our derived answer: $7$

So they agree.