Hyperbola Tangent Geometry and Greatest Integer Value

To find the greatest integer less than or equal to $\frac{\beta\delta}{9} \sin \frac{\alpha}{2}$, we analyze the geometry of the given hyperbola and its focal properties.

1. Hyperbola Parameters and Foci

The equation of the hyperbola is given by: $\frac{x^2}{100} - \frac{y^2}{64} = 1$

Here, $a^2 = 100 \implies a = 10$ and $b^2 = 64 \implies b = 8$.

The eccentricity $e$ is calculated as: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{64}{100}} = \sqrt{\frac{164}{100}} = \frac{\sqrt{164}}{10}$

The distance between the two foci $S$ and $S_1$ is: $S_1 S = 2ae = 2 \times 10 \times \frac{\sqrt{164}}{10} = 2\sqrt{164} = 4\sqrt{41}$

Thus, $(S_1 S)^2 = (4\sqrt{41})^2 = 656$.

2. Tangent and Angle Bisector Property

Let $P$ be a point on the hyperbola in the first quadrant. By the reflection property of hyperbolas, the tangent at $P$ is the interior angle bisector of $\angle S_1 P S = \alpha$.

Let line $L$ pass through focus $S$ and be parallel to the tangent at $P$. Since line $L$ is parallel to the interior angle bisector of $\angle S_1 P S$, and intersects the line $S_1 P$ at $P_1$, we consider the triangle $\triangle P P_1 S$:

• The angle bisector at $P$ divides $\angle S_1 P S$ into two equal angles of $\frac{\alpha}{2}$.
• Since $S P_1$ is parallel to the angle bisector, by alternate interior angles, $\angle P S P_1 = \frac{\alpha}{2}$.
• By corresponding angles, $\angle P P_1 S = \frac{\alpha}{2}$.

Since $\angle P S P_1 = \angle P P_1 S = \frac{\alpha}{2}$, $\triangle P P_1 S$ is an isosceles triangle with: $P P_1 = PS$

3. Lengths and Distance Calculations

Let $S_1 P = \beta$. By the definition of a hyperbola, the difference of focal distances is constant and equal to $2a$: $S_1 P - SP = 2a = 2(10) = 20 \implies SP = \beta - 20$

Since $P_1$ lies on the line segment $S_1 P$ and $P P_1 = SP = \beta - 20$, the segment $S_1 P_1$ is: $S_1 P_1 = S_1 P - P P_1 = \beta - (\beta - 20) = 20$

In the isosceles triangle $\triangle P P_1 S$ with equal sides $P P_1 = PS = \beta - 20$ and base angles equal to $\frac{\alpha}{2}$, the perpendicular distance $\delta$ from $P$ to the line $S P_1$ is given by: $\delta = P P_1 \sin \frac{\alpha}{2} = (\beta - 20) \sin \frac{\alpha}{2}$

4. Applying the Law of Cosines on $\triangle S_1 P S$

In $\triangle S_1 P S$, the sides are $S_1 P = \beta$, $SP = \beta - 20$, and $S_1 S = 4\sqrt{41}$, with angle $\angle S_1 P S = \alpha$. By the Law of Cosines: $(S_1 S)^2 = (S_1 P)^2 + (SP)^2 - 2(S_1 P)(SP) \cos \alpha$

Substituting the known values: $656 = \beta^2 + (\beta - 20)^2 - 2\beta(\beta - 20) \cos \alpha$ $656 = \beta^2 + \beta^2 - 40\beta + 400 - 2\beta(\beta - 20) \cos \alpha$ $256 = 2\beta^2 - 40\beta - 2\beta(\beta - 20) \cos \alpha$ $128 = \beta(\beta - 20)(1 - \cos \alpha)$

Using the half-angle identity $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$: $128 = 2 \beta (\beta - 20) \sin^2 \frac{\alpha}{2} \implies \beta(\beta - 20) \sin^2 \frac{\alpha}{2} = 64$

5. Final Evaluation

Now, we evaluate the required expression: $E = \frac{\beta\delta}{9} \sin \frac{\alpha}{2}$

Substitute $\delta = (\beta - 20) \sin \frac{\alpha}{2}$: $E = \frac{\beta(\beta - 20) \sin^2 \frac{\alpha}{2}}{9} = \frac{64}{9} \approx 7.111$

The greatest integer less than or equal to $E$ is: $\left\lfloor \frac{64}{9} \right\rfloor = 7$