1. Identify the parabola and its key points

   The parabola is $y^2=8x$ Comparing with the standard form $y^2=4ax$ we get $4a=8 \implies a=2.$

   So the parabola is $y^2=4ax \quad \text{with } a=2.$

2. Find the endpoints of the latus rectum

   For the parabola $y^2=4ax$, the endpoints of the latus rectum are $(a,2a) \quad \text{and} \quad (a,-2a).$

   Since $a=2$, these are $(2,4) \quad \text{and} \quad (2,-4).$

3. Given point $P$ on the parabola

   The point is $P\left(\frac12,2\right).$

   Check: $y^2=2^2=4, \qquad 8x=8\cdot \frac12=4,$ so $P$ lies on the parabola.

Part A: Compute $\Delta_1$

4. Triangle formed by the latus rectum endpoints and $P$

   The three points are $A=(2,4), \quad B=(2,-4), \quad P=\left(\frac12,2\right).$

   Segment $AB$ is vertical with length $AB=4-(-4)=8.$

   The perpendicular distance of $P$ from the line $x=2$ is $\left|2-\frac12\right|=\frac32.$

   Therefore, $\Delta_1=\frac12\times 8 \times \frac32=6.$

Part B: Compute $\Delta_2$

5. Equation of tangent to $y^2=4ax$ at point $(x_1,y_1)$ on the parabola

   Standard tangent form is $yy_1=2a(x+x_1).$

   Here $a=2$, so $yy_1=4(x+x_1).$

6. Tangents at the latus rectum endpoints

   • At $(2,4)$: $4y=4(x+2) \implies y=x+2.$

   • At $(2,-4)$: $-4y=4(x+2) \implies y=-x-2.$

7. Tangent at $P\left(\frac12,2\right)$

   $2y=4\left(x+\frac12\right)$ $2y=4x+2$ $y=2x+1.$

8. Find the triangle formed by these three tangents

   The three lines are $L_1: y=x+2,$ $L_2: y=-x-2,$ $L_3: y=2x+1.$

   Their pairwise intersections are:

   • $L_1$ and $L_2$: $x+2=-x-2 \implies 2x=-4 \implies x=-2,$ $y=0.$ So point is $Q=(-2,0).$

   • $L_1$ and $L_3$: $x+2=2x+1 \implies x=1,$ $y=3.$ So point is $R=(1,3).$

   • $L_2$ and $L_3$: $-x-2=2x+1 \implies -3x=3 \implies x=-1,$ $y=-1.$ So point is $S=(-1,-1).$

9. Area of triangle $QRS$

   Using the coordinate area formula: $\Delta_2=\frac12 \left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$

   Substituting $Q=(-2,0),\ R=(1,3),\ S=(-1,-1),$

   $\Delta_2=\frac12\left[(-2)(3-(-1))+1((-1)-0)+(-1)(0-3)\right]_{\text{abs}}$ $=\frac12\left[(-2)(4)+1(-1)+(-1)(-3)\right]_{\text{abs}}$ $=\frac12\left[-8-1+3\right]_{\text{abs}}$ $=\frac12\left|-6\right|=3.$

Part C: Ratio

10. Therefore, $\frac{\Delta_1}{\Delta_2}=\frac{6}{3}=2.$

So the required integer answer is $\boxed{2}$

The derived answer matches the stored correct answer.