A tangent to the ellipse x²/9 + y²/2 = 1 intersects the circle x² + y² = 9 at the points P and Q. It is known that R is a point on the circle x²+ y² = 9. Each of P, Q and R are located above the x-axis.

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 A-Z
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 2008-03-30 16:48:15