1. Form the equations of the two parabolas

Both parabolas have focus $F(4,3)$.

Parabola 1: directrix is the $x$-axis

Directrix: $y=0$

For any point $P(x,y)$ on the parabola, $PF = \text{distance from }P\text{ to directrix}.$ So, $\sqrt{(x-4)^2+(y-3)^2}=|y|.$ Squaring, $(x-4)^2+(y-3)^2=y^2.$ Expanding, $x^2-8x+16+y^2-6y+9=y^2$ $x^2-8x-6y+25=0.$ So the first parabola is $x^2-8x-6y+25=0.$

Parabola 2: directrix is the $y$-axis

Directrix: $x=0$

Again, for any point $P(x,y)$, $\sqrt{(x-4)^2+(y-3)^2}=|x|.$ Squaring, $(x-4)^2+(y-3)^2=x^2.$ Expanding, $x^2-8x+16+y^2-6y+9=x^2$ $y^2-6y-8x+25=0.$ So the second parabola is $y^2-6y-8x+25=0.$

2. Find their intersection points

We solve $x^2-8x-6y+25=0 \quad ...(1)$ $y^2-6y-8x+25=0 \quad ...(2)$

Subtract (2) from (1): $x^2-y^2=0$ $(x-y)(x+y)=0.$ Thus intersection points satisfy either $x=y \quad \text{or} \quad x=-y.$

Case 1: $x=y$

Substitute into (1): $x^2-8x-6x+25=0$ $x^2-14x+25=0.$ Solving, $x=\frac{14\pm\sqrt{196-100}}{2}=\frac{14\pm\sqrt{96}}{2}=7\pm 2\sqrt{6}.$ So points are $A(7+2\sqrt6,\,7+2\sqrt6), \quad B(7-2\sqrt6,\,7-2\sqrt6).$

Case 2: $x=-y$

Substitute into (1): $x^2-8x+6x+25=0$ $x^2-2x+25=0.$ Discriminant: $(-2)^2-4\cdot1\cdot25=4-100=-96<0.$ So there are no real intersection points from this case.

Hence the real intersection points are only the two points found in Case 1.

3. Compute $AB^2$

The two points are $A(7+2\sqrt6,\,7+2\sqrt6), \quad B(7-2\sqrt6,\,7-2\sqrt6).$

Difference in coordinates: $\Delta x=(7+2\sqrt6)-(7-2\sqrt6)=4\sqrt6,$ $\Delta y=(7+2\sqrt6)-(7-2\sqrt6)=4\sqrt6.$

Therefore, $AB^2=(\Delta x)^2+(\Delta y)^2$ $= (4\sqrt6)^2+(4\sqrt6)^2$ $= 96+96=192.$

So, $\boxed{AB^2=192}.$

4. Check with options

Option D is $\boxed{192}.$ So the correct option is D.

5. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.