posted by Anonymous .
Find the x-coordinate of the point P on the parabola y=1-x^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,1-a^2)
dy/dx = -2x, so at our point P the slope of the tangent is -2a
equation of tangent:
y - (1-a^2) = -2a(x-a)
y - 1 + a^2 = -2ax +2a^2
2ax + y = a^2 + 1
the base of the triangle is the x-intercept of this line,
the height of the triangle is the y-intercept of this line.
x-intercept: x = (a^2 + 1)/(2a)
y-intercept: y = a^2 + 1
Area of triangle
= (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)