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?
IF the distance is minimized, then on each end of that distance line will be two curves with the same slope.
take the derivatives of each lines, set in each dy/dx equal to each other, and see if you can get a single equation which reduces.
1P) 2x+2yy'-136+12y'=0
2y'{y+6)=-2x+136
2Q) 2(y+6)y'=2x^2-256x-417
well, those two can be set equal
2x^2-256x-417=-2x+136
2x^2-254x-553=0
and you can solve for x which satisfy this. It wont take you long to see if this leads you to the two points, P and Q. I think I would give it a shot.
I have been stewing over this question since last night, even to the point of dreaming about it, how bad is that?
I agree with Bob's assessment that at the shortest line, the tangents' slopes at these two endpoints must be equal.
I equated the two derivative expressions, which are
(3x^2 - 232x - 417)/(2y+12) and
(136-2x)/(2y+12)
The denominators drop out leaving you with
3x^2 - 230x - 553 = 0
(Bob had a typo in this equation)
This led to a very simple solution of x=79 and x = -7/3
making us believe that we got something there.
subbing x=79 into the second equation even gave me a nice y = 54
but wait ....
We can't equate these two, since the (x,y)'s have different equations, and they don't intersect.
More to come ...
Ok, the first equation is a circle.
I completed the square and wrote it as
(x-68)^2 + (y+6)^2 = 100
centre is (68,-6) and radius is 10
The line of shortest distance must run through the centre of that circle.
So let Q(x,y) be the point of contact on the second curve.
So the slope of the line of shortest distance is (y+6)/(x-68)
and the slope of the tangent at Q for the second curve is (3x^2 - 232x - 417)/(2y+12)
But they must be negative reciprocals of each other, so ... (skipping the algebra) ...
3x^2 - 230x - 553 = 0 which is the same equation we had in my previous reply
that gave us x = 79, y = 54
subbing (79,54) into the slope of the line running through the centre gives me
60/11
subbing (79,54) into the derivative of the second equation,(the slope of the tangent), gives me -22/120
YEAHHH!
So now to find P , solve
y+6 = (60/11)(x-68) with the circle equation.
To find the actual shortest distance we don't even need point P.
I would simply be the distance form (79,54) to the centre(68,-6) less the radius of 10
Shortest distance
= √((79-68)^2 + (54+6)^2) - 10
= √3721 - 10
= 61 - 10
= 51
Wow, what a question.
Last revision.
Back in x= 79, subbing back into the second equation gave me y = 54
but there was also y = -66
Calculating THAT distance gives me
√(79-68)^2 + (-66+6)^2) - 10
= 51 ..... the same as before!!
I actually ran the two curves through my graphing program (in old 1970's DOS program written in BASIC)
Somebody must have gone to great length to make up this question. The circle lies on the line of symmetry of the second equation, notice (54+(-66))/2 = -6.
To find the equation containing x and y after the minimization has occurred, we need to set up the function that represents the distance between point P and point Q.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this problem, point P can be represented by the equation x^2 + y^2 - 136x + 12y + 4560 = 0, and point Q can be represented by the equation (y + 6)^2 = x^3 - 116x^2 - 417x + 267460.
We need to minimize the distance d, which is calculated using the above formula.
Let's calculate the distance d in terms of (x) and (y):
d = sqrt((x - P_x)^2 + (y - P_y)^2)
Substituting the equations of P and Q, we get:
d = sqrt((x - x₁)² + (y - y₁)²)
= sqrt((x - x₁)² + (y + 6 - x₁^3 + 116x₁² + 417x - 267460)²)
To find the minimum distance, we can differentiate this function with respect to x and y and set both derivatives equal to zero.
By minimizing the distance function, we can find the values of x and y that give the smallest distance between P and Q.