Consider the hyperbola $$\dfrac{x^2}{100} - \dfrac{y^2}{64} = 1$$ with foci at S and S$$_1$$, where S lies on the positive x-axis. Let P be a point on the hyperbola, in the first quadrant. Let $$\angle$$SPS$$_1 = \alpha$$, with $$\alpha < \dfrac{\pi}{2}$$. The straight line passing through the point S and having the same slope as that of the tangent at P to the hyperbola, intersects the straight line S$$_1$$P at P$$_1$$. Let $$\delta$$ be the distance of P from the straight line SP$$_1$$, and $$\beta = S_1P$$. Then the greatest integer less than or equal to $$\dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2}$$ is _______.

The given hyperbola is $$\dfrac{x^{2}}{100}-\dfrac{y^{2}}{64}=1$$.

For this hyperbola  • semi-transverse axis $$a=10$$,   • semi-conjugate axis $$b=8$$,  • eccentricity $$e=\sqrt{1+\dfrac{b^{2}}{a^{2}}}= \sqrt{1+\dfrac{64}{100}}=\dfrac{\sqrt{41}}{5}$$.

Hence the foci are $$S\,(ae,0)=\left(2\sqrt{41},0\right),\qquad S_{1}\,(-ae,0)=\left(-2\sqrt{41},0\right).$$

Let the point $$P(x,y)$$ lie in the first quadrant on the right branch of the hyperbola. Because $$x^{2}/100-y^{2}/64=1$$ and $$x\gt 0,\;y\gt 0$$ we may treat $$x$$ as the only free variable (with $$x\gt 10$$) and write

$$y=\dfrac45\sqrt{x^{2}-100}\qquad\bigl(y\gt 0\bigr).$$

1. Slope of the tangent at P
Differentiate $$\dfrac{x^{2}}{100}-\dfrac{y^{2}}{64}=1:$$ $$\dfrac{2x}{100}-\dfrac{2y}{64}\dfrac{dy}{dx}=0 \;\Longrightarrow\; \dfrac{dy}{dx}= \dfrac{64}{100}\dfrac{x}{y}= \dfrac{16x}{25y}.$$ Hence the tangent slope is $$m=\dfrac{16x}{25y}.$$

2. Line through S having the same slope
Through $$S(2\sqrt{41},0)$$ with slope $$m$$ the line is $$y=m\bigl(x-2\sqrt{41}\bigr) \;\Longrightarrow\; mx-y-2m\sqrt{41}=0.\tag{-1}$$

3. Perpendicular distance of P from this line
Using the distance formula,

$$\delta=\dfrac{\left|mx-y-2m\sqrt{41}\right|}{\sqrt{m^{2}+1}}.$$ Compute the numerator first: $$$ \begin{aligned} mx-y&=\dfrac{16x}{25y}\,x-y =\dfrac{16x^{2}-25y^{2}}{25y}\\ &=\dfrac{16x^{2}-25\Bigl(\dfrac{16}{25}(x^{2}-100)\Bigr)}{25y} =\dfrac{16x^{2}-16x^{2}+1600}{25y} =\dfrac{64}{y}. \end{aligned} $$$ So $$$ mx-y-2m\sqrt{41}= \dfrac{64}{y}- 2\Bigl(\dfrac{16x}{25y}\Bigr)\sqrt{41} =\dfrac{1}{y}\Bigl[64-\dfrac{32x\sqrt{41}}{25}\Bigr]. $$$ Since $$x\gt 10$$ the bracket is negative, and its absolute value is $$$ \left|\;64-\dfrac{32x\sqrt{41}}{25}\right| =\dfrac{32x\sqrt{41}-1600}{25}. $$$ Next, $$$ m^{2}+1= \Bigl(\dfrac{16x}{25y}\Bigr)^{2}+1 =\dfrac{256x^{2}+625y^{2}}{625y^{2}} =\dfrac{16\bigl(41x^{2}-2500\bigr)}{625y^{2}}, $$$ hence $$\sqrt{m^{2}+1}= \dfrac{4\sqrt{41x^{2}-2500}}{25y}.$$ Putting everything together, $$$ \delta=\dfrac{ \dfrac{32x\sqrt{41}-1600}{25y}} { \dfrac{4\sqrt{41x^{2}-2500}}{25y}} =\dfrac{32x\sqrt{41}-1600}{4\sqrt{41x^{2}-2500}} =\dfrac{8x\sqrt{41}-400}{\sqrt{41x^{2}-2500}}.\tag{-2} $$$

4. Distance $$\beta=S_{1}P$$
$$$ \beta^{2}= (x+2\sqrt{41})^{2}+y^{2} =x^{2}+4x\sqrt{41}+164+\dfrac{16}{25}(x^{2}-100) =\dfrac{41}{25}x^{2}+4x\sqrt{41}+100.\tag{-3} $$$

5. The angle $$\alpha=\angle SPS_{1}$$
In triangle $$\triangle SS_{1}P$$ let $$p=SP,\quad q=S_{1}P=\beta,\quad d=SS_{1}=4\sqrt{41}.$$ For a point on the right branch of a hyperbola, $$p-q=2a=20\;\Longrightarrow\; p=q+20.\tag{-4}$$ Using the well-known half-angle identity for any triangle $$\sin\dfrac{\alpha}{2}=\sqrt{\dfrac{(s-p)(s-q)}{pq}},$$ where $$s=\dfrac{p+q+d}{2}$$ is the semiperimeter. Substituting $$p-q=20$$ gives $$$ (s-p)(s-q)=\dfrac{d-20}{2}\cdot\dfrac{d+20}{2} =\dfrac{d^{2}-400}{4} =\dfrac{256}{4}=64, $$$ because $$d^{2}=(4\sqrt{41})^{2}=656.$$ Thus $$$ \sin\dfrac{\alpha}{2}= \sqrt{\dfrac{64}{pq}} =\dfrac{8}{\sqrt{pq}}.\tag{-5} $$$

6. Required expression
$$$ E=\dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2} =\dfrac{\beta\delta}{9}\cdot\dfrac{8}{\sqrt{p\beta}} =\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{p}} =\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{\beta+20}}\quad\bigl(\text{by }( -4)\bigr).\tag{-6} $$$

7. Behaviour of $$E$$ as $$x$$ varies
From $$( -2)$$, $$\delta(x)=\dfrac{8x\sqrt{41}-400}{\sqrt{41x^{2}-2500}}.$$ A simple derivative test (or numerical check) shows that $$\delta(x)$$ is increasing and $$$ \lim_{x\to\infty}\delta(x)=8\quad\text{while}\quad \delta(x)\lt 8\ \text{for every finite }x.  $$$ Likewise, from $$( -3)$$ we have $$\beta(x)\to\infty$$ as $$x\to\infty$$, so $$$ 0\lt \sqrt{\dfrac{\beta}{\beta+20}}\lt 1,\qquad \lim_{x\to\infty}\sqrt{\dfrac{\beta}{\beta+20}}=1. $$$ Therefore, taking the limit in $$( -6)$$, $$$ \boxed{\; \lim_{x\to\infty}E=\dfrac{8}{9}\times 8= \dfrac{64}{9}=7.111\ldots\; }. $$$ For every finite $$x$$ we have $$\delta\lt 8$$ and $$\sqrt{\beta/(\beta+20)}\lt 1$$, hence $$$ E=\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{\beta+20}} \lt \dfrac{8}{9}\times 8=\dfrac{64}{9}. $$$ Consequently, $$$ 7 \lt E \lt \dfrac{64}{9}\;(=7.111\ldots). $$$ Thus $$E$$ can be made arbitrarily close to $$64/9$$ from below but never reaches or exceeds it. The greatest integer $$\le E$$ for every allowable point $$P$$ is therefore $$7$$.

Answer: 7