Calculus
posted by Ivan .
find the xcoordinate of the point P on the parabola y=1x^2 for 0<x<1 where the triangle that is enclosed by the tangent line at P and the coordinate axes has the smallest area.

let the point of contact in the first quadrant be P(a,b) or P(a,1a^2)
slope of the tangent is 2x (the derivative)
so at P the slope is 2a
equation of the tangent is
(y  (1a^2))/(xa) = 2a
which simplifies to
2ax + y = a^2 + 1
the base of the triangle will be the xintercept of the tangent
the height of the triangle will be the yintercept of the tangent.
xintercept : let y = 0, then x = (a^2+1)/(2a)
yintercept : 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.
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