Posted by justin on Tuesday, October 16, 2007 at 10:10pm.
let p(a,c/a) be the point on the hyperbola
for xy=c
dy/dx = -y/x, so at P the slope = -c/a^2
equation of tangent line:
y - c/a = -c/a^2(x - a) which when simplified is
cx - a^2y=-2ac
for x-intercept, let y=0, then x = 2a
for y-intercept, let x=0, then y = 2c/a
1. take the midpoint of (0,2c/a) and (2a,0) and what do you get????
2. aren't your x and y intercepts the base and height of your triangle??
take area = 1/2 base*height
= .....
= c which is the constant of the original equation!!!
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