Determine the equation of k in the form of k(x) = ax2 if B is equidistant from A to C.

Let B be $(x_B, y_B)$. A is $(6,0)$ and C is $(0,-6)$.
Since B is equidistant from A and C, $BA^2 = BC^2$.

$$(x_B - 6)^2 + (y_B - 0)^2 = (x_B - 0)^2 + (y_B - (-6))^2$$ $$x_B^2 - 12x_B + 36 + y_B^2 = x_B^2 + y_B^2 + 12y_B + 36$$ $$-12x_B = 12y_B$$ $$y_B = -x_B$$

From 4.3, the equation of line p is $y = x - 6$.
Substitute $y_B = -x_B$ into the line equation:

$$-x_B = x_B - 6$$ $$-2x_B = -6$$ $$x_B = 3$$

Now find $y_B$:

$$y_B = -x_B = -3$$

So, the coordinates of B are $(3, -3)$.

Point B(3, -3) lies on the parabola $k(x) = ax^2$. Substitute $x=3$ and $k(x)=-3$:

$$-3 = a(3)^2$$ $$-3 = 9a$$ $$a = \frac{-3}{9}$$ $$a = -\frac{1}{3}$$

Substitute the value of $a$ back into $k(x) = ax^2$:

$$k(x) = -\frac{1}{3}x^2$$