Posted by **Anonymous** on Tuesday, November 24, 2009 at 5:55pm.

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.

- calculus -
**Reiny**, Tuesday, November 24, 2009 at 8:23pm
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

= (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)

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