A tangent to the ellipse x
^{2}/9 + y
^{2}/2 = 1 intersects the circle x
^{2} + y
^{2} = 9 at the points P and Q. It is known that R is a point on the circle x
^{2}+ y
^{2} = 9. Each of P, Q and R are located
above the
xaxis.
For example, the coordinates of R cannot be (3,0), since (3,0) is not located above the x axis.
Determine the maximum area of the triangle PQR.
(In reply to
Well..... by Charlie)
DEFDBL AZ
CLS
minE = .1#: maxE = SQR(2#)
minC = 0#: maxC = 3#
FOR ey0 = minE TO maxE STEP (maxE  minE) / 6.000001#
ct = ct + 1
LOCATE 1, INT(ct * 10)
PRINT USING "##.######"; ey0;
NEXT
ctc = 0
FOR cy0 = minC TO maxC STEP (maxC  minC) / 40
ctc = ctc + 1
cx0 = SQR(9  cy0 * cy0)
LOCATE ctc + 1, 1
PRINT USING "#.####"; cy0
ct = 0
FOR ey0 = minE TO maxE STEP (maxE  minE) / 6.000001#
ex0 = SQR(9 * (1  ey0 * ey0 / 2))
slope = 1 / (2 * SQR(2  2 * ex0 * ex0 / 9))
b = ey0  slope * ex0
m = slope
x1 = (2 * b * m  SQR(4 * b * b * m * m  4 * (m * m + 1) * (b * b  9))) / (2 * (m * m + 1))
x2 = (2 * b * m + SQR(4 * b * b * m * m  4 * (m * m + 1) * (b * b  9))) / (2 * (m * m + 1))
y1 = m * x1 + b
y2 = m * x2 + b
a = SQR((x1  x2) ^ 2 + (y1  y2) ^ 2)
b = SQR((x1  cx0) ^ 2 + (y1  cy0) ^ 2)
c = SQR((x2  cx0) ^ 2 + (y2  cy0) ^ 2)
s = (a + b + c) / 2
area = SQR(s * (s  a) * (s  b) * (s  c))
ct = ct + 1
LOCATE ctc + 1, INT(ct * 10)
PRINT USING "#.#######"; area
NEXT
NEXT

Posted by Charlie
on 20080330 16:48:15 