P is a point on the parabola f(x) = 1-x^2. A tangent is drawn at point P cuts x axes and y axes to form a triangle. Find the coordinates of P that will minimize the area of the triangle

2. Hello, pita!

$P$ is a point on the parabola: $f(x) \:= \:1-x^2$
A tangent is drawn at point $P$ cuts x-axis and y-axis to form a triangle.
Find the coordinates of $P$ that will minimize the area of the triangle.

The point P has coordinates: . $\left(p,\:1-p^2\right)$

The derivative is: . $f'(x) \:=\:-2x$
The tangent at $P$ has slope: . $m \:=\:-2p$

The tangent at $P$ has the equation: . $y - (1-p^2) \;=\;-2p(x-p)$
. . which simplifies to: . $y \;=\;-2px + p^2+1$

$\text{It has }x\text{-intercept }\left(\frac{p^2+1}{2p},\:0\right)\;\text{ and }\;y\text{-intercept }\left(0,\:p^2+1\right)$

The area of the triangle is: . $A \;=\;\frac{1}{2}\left(\frac{p^2+1}{2p}\right)\left (p^2+1\right) \;=\;\frac{1}{4}\left(\frac{p^4 + 2p^2 + 1}{p}\right)$

Therefore: . $A \;=\;\frac{1}{4}\left(p^3 + 2p + p^{-1}\right)$ .is the function to be maximized.