There is a point (P) on the graph of [x^2+y^2- 136 x + 12 y + 4560 = 0] and a point(Q) on the graph of

[(y + 6)^2 = x^3 - 116 x^2 - 417 x + 267460] such that the distance between them is as small as possible.

To solve this problem, we let ((x,y) be the coordinates of the point Q. Then we need to minimize the following function of (x) and (y): What will be the equation containing x and y after minimization has occured?

To find the equation containing x and y after minimization has occurred, we first need to define the function we want to minimize. In this case, the function is the distance between points P and Q, given by the distance formula:

distance = sqrt((x1 - x2)^2 + (y1 - y2)^2)

Recall that point P lies on the graph of x^2 + y^2 - 136x + 12y + 4560 = 0, which we can rewrite as:

(x - 68)^2 + (y + 6)^2 = 4570

Hence, we have the coordinates of point P as (68, -6).

Next, point Q lies on the graph of (y + 6)^2 = x^3 - 116x^2 - 417x + 267460. We can substitute the values of x and y into the distance formula to get the distance in terms of x:

distance = sqrt((x - 68)^2 + ((y + 6) - y)^2)

Simplifying this further:

distance = sqrt((x - 68)^2 + 36)

Now, we need to minimize this distance function. To find the minimum, we can take the derivative of the distance with respect to x and set it to zero, since the minimum occurs when the slope of the distance function is zero.

Differentiating the distance function with respect to x:

d(distance)/dx = [(x - 68) / sqrt((x - 68)^2 + 36)] * 2(x - 68)

Setting this equal to zero:

[(x - 68) / sqrt((x - 68)^2 + 36)] * 2(x - 68) = 0

Considering the factors separately, we have:

x - 68 = 0 (Case 1)
(x - 68) / sqrt((x - 68)^2 + 36) = 0 (Case 2)

In Case 1, solving for x gives: x = 68.

In Case 2, solving for x gives: (x - 68) = 0, which also leads to x = 68.

Therefore, we have x = 68 as the x-coordinate for the minimum distance. Substituting this value back into the equation containing y, we can solve for y:

(y + 6)^2 = 68^3 - 116(68)^2 - 417(68) + 267460

Simplifying this equation will yield a quadratic equation in terms of y, which can be further solved to obtain the equation containing x and y after minimization has occurred.