A vertical line passing through the point (h, 0) intersects the ellipse $x^\frac{2}{4} + y^\frac{2}{3} = 1$ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If $\Delta(h) =$ area of the triangle PQR$; \Delta_{1} = \max$ over $\frac{1}{2} \le h \le 1$ of $\Delta(h)$ and $\Delta_{2} = \min$ over $\frac{1}{2} \le h \le 1$ of $\Delta(h),$ then $(8/sqrt5) \Delta_{1} - 8 \Delta_{2} =$ ____.

$\frac{x^2}{4} + \frac{y^2}{3} = 1$ $\text{line } x=h \Rightarrow P, Q;\quad \text{tangents at } P, Q \text{ meet at } R$ $\Delta(h) = [\triangle PQR],\quad h \in \left[\frac12, 1\right]$ $\text{find } \Delta_1 = \max \Delta,\qquad \Delta_2 = \min \Delta$ $P(h,k),\quad Q(h,-k),\quad k = \frac{\sqrt3}{2}\sqrt{4-h^2}$ $T_P:\ \frac{hx}{4} + \frac{ky}{3} = 1$ $T_Q:\ \frac{hx}{4} - \frac{ky}{3} = 1$ $R = \left(\frac{4}{h},\ 0\right)$ $\Delta(h) = \frac{1}{2}\cdot(PQ)\cdot\text{height} = \frac{1}{2}(2k)\left(\frac{4}{h}-h\right)$ $= k\left(\frac{4}{h}-h\right)$ $\Delta(h) = \frac{\sqrt3}{2}\cdot\sqrt{4-h^2}\cdot\frac{4-h^2}{h}$ $= \frac{\sqrt3}{2}\cdot\frac{(4-h^2)^{3/2}}{h}$ $\frac{d}{dh}\ln\Delta = \frac{3}{2}\cdot\frac{-2h}{4-h^2} - \frac{1}{h} = 0 \Rightarrow h^2 = -2$ $\Rightarrow \text{no critical point} \Rightarrow \Delta \text{ decreasing on } \left[\tfrac12,1\right]$ $\Delta_1 = \Delta\!\left(\tfrac12\right) = \frac{\sqrt3}{2}\cdot\frac{(15/4)^{3/2}}{1/2}$ $= \frac{45\sqrt5}{8}$ $\Delta_2 = \Delta(1) = \frac{\sqrt3}{2}\cdot\frac{3^{3/2}}{1} = \frac{9}{2}$ $\frac{8}{\sqrt5}\Delta_1 - 8\Delta_2 = \frac{8}{\sqrt5}\cdot\frac{45\sqrt5}{8} - 8\cdot\frac{9}{2}$ $= 45 - 36 = 9$