To solve the problem step by step, we will find the area of triangle PQR formed by the tangents to the ellipse at points P and Q, and then compute the required expression.
### Step 1: Identify the points of intersection
The vertical line through the point (h, 0) intersects the ellipse given by the equation:
\[
\frac{x^2}{4} + \frac{y^2}{3} = 1
\]
Substituting \(x = h\) into the ellipse equation, we can find the corresponding \(y\) values:
\[
\frac{h^2}{4} + \frac{y^2}{3} = 1
\]
Rearranging gives:
\[
\frac{y^2}{3} = 1 - \frac{h^2}{4}
\]
\[
y^2 = 3\left(1 - \frac{h^2}{4}\right)
\]
\[
y = \pm \sqrt{3\left(1 - \frac{h^2}{4}\right)}
\]
Thus, the points of intersection P and Q are:
\[
P\left(h, \sqrt{3\left(1 - \frac{h^2}{4}\right)}\right), \quad Q\left(h, -\sqrt{3\left(1 - \frac{h^2}{4}\right)}\right)
\]
### Step 2: Find the equations of the tangents at points P and Q
The equation of the tangent to the ellipse at point \((x_1, y_1)\) is given by:
\[
\frac{xx_1}{4} + \frac{yy_1}{3} = 1
\]
For point P:
\[
\frac{x h}{4} + \frac{y \sqrt{3\left(1 - \frac{h^2}{4}\right)}}{3} = 1
\]
For point Q (similar equation):
\[
\frac{x h}{4} - \frac{y \sqrt{3\left(1 - \frac{h^2}{4}\right)}}{3} = 1
\]
### Step 3: Find the intersection point R of the tangents
To find the coordinates of point R, we solve the two tangent equations simultaneously. Adding the two equations gives:
\[
\frac{2xh}{4} = 2 \Rightarrow x = \frac{4}{h}
\]
Substituting \(x = \frac{4}{h}\) into one of the tangent equations to find \(y\):
\[
\frac{\frac{4}{h} h}{4} + \frac{y \sqrt{3\left(1 - \frac{h^2}{4}\right)}}{3} = 1
\]
This simplifies to:
\[
1 + \frac{y \sqrt{3\left(1 - \frac{h^2}{4}\right)}}{3} = 1
\]
Thus,
\[
\frac{y \sqrt{3\left(1 - \frac{h^2}{4}\right)}}{3} = 0 \Rightarrow y = 0
\]
So, the coordinates of point R are:
\[
R\left(\frac{4}{h}, 0\right)
\]
### Step 4: Calculate the area of triangle PQR
The area of triangle PQR can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base PQ is the vertical distance between P and Q:
\[
PQ = 2\sqrt{3\left(1 - \frac{h^2}{4}\right)}
\]
The height from R to the line segment PQ is the horizontal distance from R to the line \(x = h\):
\[
\text{Height} = \left|\frac{4}{h} - h\right| = \left|\frac{4 - h^2}{h}\right|
\]
Thus, the area of triangle PQR is:
\[
\Delta(h) = \frac{1}{2} \times 2\sqrt{3\left(1 - \frac{h^2}{4}\right)} \times \left|\frac{4 - h^2}{h}\right| = \frac{4\sqrt{3\left(1 - \frac{h^2}{4}\right)(4 - h^2)}}{2h}
\]
### Step 5: Maximize the area
To find \(\Delta_1\) and \(\Delta_2\), we need to maximize \(\Delta(h)\) over the interval \(\left[\frac{1}{2}, 1\right]\).
After finding the maximum area \(\Delta_1\) at \(h = \frac{1}{2}\) and minimum area \(\Delta_2\) at \(h = 1\), we compute:
\[
\Delta_1 = \frac{41\sqrt{5}}{8}, \quad \Delta_2 = \frac{9}{2}
\]
### Step 6: Calculate the final expression
Now we compute:
\[
\frac{8}{\sqrt{5}} \Delta_1 - 8 \Delta_2 = \frac{8}{\sqrt{5}} \cdot \frac{41\sqrt{5}}{8} - 8 \cdot \frac{9}{2}
\]
This simplifies to:
\[
41 - 36 = 5
\]
Thus, the final answer is:
\[
\boxed{5}
\]
