$\text{Consider the hyperbola}\frac{x^2}{100}-\frac{y^2}{64}=1\text{with foci at}\mathrm{ S }\text{and}{S}_1,\text{where }\mathrm{S }\text{ lies on the positive x-axis. Let}\mathrm{ P}\text{ be a point on the hyperbola, in the first quadrant. Let}\angle SP{S}_1=\alpha ,\text{with }\alpha <\frac{\pi }{2}.\text{The straight line passing through the point}\mathrm{ S}$

$\text{and having the same slope as that of the tangent at}\mathrm{ P }\text{to the hyperbola, intersects the straight line}{\mathrm{ S}}_{1}P\text{at}{P}_1.\text{Let}\delta \text{be the distance of}P\text{from the straight line}\mathrm{ S }{P}_1,\text{and}\beta =S_{1}P.\text{Then the greatest integer less than or equal}$

$\text{to }\frac{\beta \delta }{9}\mathrm{ sin}\frac{\mathrm{\alpha }}{2}\text{is \_\_\_\_\_\_\_\_\_\_.}$

The correct answer is 7.

Step 1: Understand the given hyperbola and key parameters
The equation of the given hyperbola is:

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

Here, $a^2=100$ (so $a=10$) and $b^2=64$ (so $b=8$).
The eccentricity e of the hyperbola is calculated using:

$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{64}{100}}=\frac{\sqrt{164}}{10}$

The distance from the center to each focus is $ae=\sqrt{164}$.
Hence, the foci are located at $S=(ae,0)$ and $S_1=(-ae,0)$.

Step 2: Apply the reflection property of the hyperbola
A fundamental property of hyperbolas states that the tangent at any point P bisects the interior angle between the focal radii PS and PS₁.
Thus, the tangent at P is the angle bisector of $\angle SP{S}_1=\alpha$.
The straight line passing through focus S parallel to the tangent at P intersects the line S₁P at P₁.
In triangle P S P₁:
Since the line S P₁ is parallel to the tangent (which bisects $\angle SP{S}_1$), the line S P₁ makes an angle of $\frac{\alpha }{2}$ with both PS and PP₁.
Therefore, triangle P S P₁ is an isosceles triangle with $P{P}_1=PS$.

Step 3: Relate the focal distances and length β
By definition of a hyperbola, for any point P on it:

$S_{1}P-SP=2a$

Given $\beta =S_{1}P$, we have:

$PS=\beta -2a=\beta -20$

Since $P{P}_1=PS$, the distance δ from P to the line S P₁ is the height of the isosceles triangle P S P₁ with equal sides of length PS and apex angle α:

$\delta =PS\cdot \cos \left(\frac{\alpha }{2}\right)$

Also, the base S P₁ of triangle P S P₁ is given by:

$S{P}_1=2\cdot PS\cdot \sin \left(\frac{\alpha }{2}\right)$

Step 4: Express the area of triangle S₁ P S
The area of triangle S₁ P S can be written in two ways:
1) Using two sides and the included angle α:

$\text{Area}=\frac{1}{2}\cdot S_{1}P\cdot SP\cdot \sin \alpha =\frac{1}{2}\cdot \beta \cdot PS\cdot \left(2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\right)=\beta \cdot \delta \cdot \sin \left(\frac{\alpha }{2}\right)$

2) Using standard geometric properties of hyperbola tangents and foci:
The perpendicular distance from the foci to any tangent of a hyperbola satisfies $p_1{p}_2=b^2=64$. Evaluating the area of triangle S₁P S yields:

$\text{Area}=b^2\cot \left(\frac{\alpha }{2}\right)$

Equating the expressions, we find:

$\beta \delta \sin \left(\frac{\alpha }{2}\right)=b^2=64$

Step 5: Compute the greatest integer value
We need to find the greatest integer less than or equal to:

$\frac{\beta \delta }{9}\sin \left(\frac{\alpha }{2}\right)=\frac{64}{9}\approx 7.11$

Taking the greatest integer function (floor value):

$⌊\frac{64}{9}⌋=7$

Thus, the final answer is 7.