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 Trapezing The Parabola (Posted on 2007-12-01)
The two diagonals PR and QS of a trapezoid PQRS intersect at the point (1, 0) and PS is parallel to QR. All the four vertices of the trapezoid lie on the parabola y2 = 4x and PR = QS = 100/9.

Determine the area of the trapezoid PQRS.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 Analytic Solution Comment 3 of 3 |

The points on the parabola y^2 = 4x can be parameterized as (x,y) = (z^2, 2z).

Let a and b be variables with b>1>a>0.  Then let the coordinates of P be (a^2,2a), let Q be (b^2,2b), let R be (b^2,-2b), and let S be (a^2,-2a). Also let C denote point (1,0).

For PC and CR to be colinear, the slopes must be equal:
(2a)/(a^2-1) = (2b)/(1-b^2)
a^2*b - b = a = a*b^2
(ab-1)*(a+b) = 0
a = 1/b

Since the length of PR is 100/9:
(100/9)^2 = (b^2-a^2)^2 + (2b+2a)^2
(10/3)^4 = b^4 - 2a^2*b^2 + a^4 + 4b^2 + 8a*b + 4a^2

Substitute a=1/b:
(10/3)^4 = b^4 - 2 + 1/b^4 + 4b^2 + 8 + 4/b^2
(10/3)^4 = (b + 1/b)^4
b = 3, a = 1/3

The trapezoid has bases of 4a and 4b and has a height of b^2-a^2.  The area is then (b^2-a^2)*(2b+2a).  Substituting b=3, a=1/3 yields an area of (9-1/9)*(6+2/3) = 59 + 7/27 = 1600/27

 Posted by Brian Smith on 2007-12-03 13:04:30

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