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 Intersection And Maximum Triangle (Posted on 2008-03-30)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 re: Well..... (the program) | Comment 2 of 4 |
(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

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