calculus
posted by Anonymous .
Find the xcoordinate of the point P on the parabola y=1x^2 (0<x=<1) where the triangle enclosed by the tangent line at p and the coordinate axes has the smallest area.

let the point of contact be P(a,1a^2)
dy/dx = 2x, so at our point P the slope of the tangent is 2a
equation of tangent:
y  (1a^2) = 2a(xa)
y  1 + a^2 = 2ax +2a^2
2ax + y = a^2 + 1
the base of the triangle is the xintercept of this line,
the height of the triangle is the yintercept of this line.
xintercept: x = (a^2 + 1)/(2a)
yintercept: y = a^2 + 1
Area of triangle
= (1/2)(a^2+1)^2/(2a)
= (a^4 + 2a^2 + 1)/a
= a^3 + 2a + 1/a
d(Area)/da = 3a^2 + 2  1/a^2
= 0 for a max/min of Area
the only real solution for the above is
a = ± 1/√3
so P is (1/√3 , 2/3)