Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then $$(AB)^{2}$$ is equal to :

We have two parabolas with the same focus $$(4,3)$$, where the directrix of the first parabola is the x-axis and the directrix of the second parabola is the y-axis. We need to find $$(AB)^2$$ where $$A$$ and $$B$$ are the intersection points.

For the first parabola with focus $$(4,3)$$ and directrix $$y=0$$, the distance from any point $$(x,y)$$ to the focus equals its distance to the directrix:

$$ \sqrt{(x-4)^2 + (y-3)^2} = |y| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = y^2 $$

which expands to

$$ (x-4)^2 + y^2 - 6y + 9 = y^2 $$

and hence

$$ (x-4)^2 = 6y - 9 \quad \cdots (1) $$

For the second parabola with focus $$(4,3)$$ and directrix $$x=0$$, we similarly have

$$ \sqrt{(x-4)^2 + (y-3)^2} = |x| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = x^2 $$

which expands to

$$ x^2 - 8x + 16 + (y-3)^2 = x^2 $$

so

$$ (y-3)^2 = 8x - 16 $$

and

$$ (y-3)^2 = 8(x - 2) \quad \cdots (2) $$

At the intersection points the distances to the focus and the two directrices are equal, implying $$|y| = |x|$$. Thus the intersection points lie on $$y = x$$ or $$y = -x$$.

In the case $$y = x$$, substituting into (1) yields

$$ (x-4)^2 = 6x - 9 $$

so

$$ x^2 - 8x + 16 = 6x - 9 $$

and

$$ x^2 - 14x + 25 = 0 $$

giving

$$ x = \frac{14 \pm \sqrt{196 - 100}}{2} = \frac{14 \pm \sqrt{96}}{2} = 7 \pm 2\sqrt{6} $$

Hence the points are $$(7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$(7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$.

In the case $$y = -x$$, substituting into (1) yields

$$ (x-4)^2 = -6x - 9 $$

so

$$ x^2 - 2x + 25 = 0 $$

The discriminant $$4 - 100 = -96 < 0$$ shows there are no real solutions.

Letting $$A = (7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$B = (7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$, we compute

$$ (AB)^2 = (4\sqrt{6})^2 + (4\sqrt{6})^2 = 96 + 96 = 192 $$

The correct answer is Option 3: 192.