Thread: area of quadrilateral
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Old June 14th, 2008, 06:22 AM
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Quote:
Originally Posted by afeasfaerw23231233 View Post
bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
C_1 : x^2 +y^2+2tx=0
C_2: x^2+y^2 +\frac {2y}t = 0
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
0,- \frac 1 t
0,0
-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}
area:
\frac 1 2 \begin{vmatrix}0 & 0 \\-t & 0 \\ 0 & -\frac 1 t \\-\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\0 & 0 \end{vmatrix}
= \frac 1 2 (\frac {t^4-1}{1+t^4})
cannot prove that the area is a constant
thanks
When t = 1, A = 1. But your answer gives A = 0 ......
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