Posted by Ivan on Thursday, November 6, 2008 at 8:27pm.
let the point of contact in the first quadrant be P(a,b) or P(a,1-a^2)
slope of the tangent is -2x (the derivative)
so at P the slope is -2a
equation of the tangent is
(y - (1-a^2))/(x-a) = -2a
which simplifies to
2ax + y = a^2 + 1
the base of the triangle will be the x-intercept of the tangent
the height of the triangle will be the y-intercept of the tangent.
x-intercept : let y = 0, then x = (a^2+1)/(2a)
y-intercept : let x = 0, then y = a^2 + 1
so the Area = (a^2 + 1)^2 /(4a)
use the quotient rule to find the derivative of Area, set that equal to zero and solve.
let me know what you got.
1
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