$(a,b)$ with $a, b \gt 0$ on $f(x)=4-x^2$ so that the area of the triangle limited by X- , Y-axis, tangent on $f(x$) at $(a,b)$ is minimal? How can I determine the point $(a,b) \in \mathbb{R^2}$ with $a, b \gt 0$ on $f(x)=4-x^2$ so that the area of the triangle which is limited by the X- and Y-axis and by the tangent on $f(x$) at $(a,b)$ is minimal? The area of the triangle is $(ycoordinate(intersection_{tangent, y-axis})*xcoordinate(intersection_{tanget,x-axis}))$/2 If we derive $f(x)$ we get $f'(x)=-2x$. Because $y=kx+d$ we get $d=f(a)+2xa = 4+x^2$ so $y(intersection_{tangent, y-axis})$ should be $4+a^2$. But somehow I cant get any further sketch original task(in german!)

let $$y=mx+n$$ the searched Tangent line at the Point $$P(a;f(a))$$ then the slope is given by $$f'(x)=-2x$$ and in $a$ $$f'(a)=-2a$$ so our Tangent line has the equation $$y=-2ax+n$$ since $$f(a)=4-a^2$$ we get $$y=-2ax+4+a^2$$ now the $y$ intercept is $$4+a^2$$ and the $x$ intercept $$x=\frac{4+a^2}{2a}$$ therefore our tringle has the area: $$A=\frac{1}{2}(4+a^2)\left(\frac{4+a^2}{2a}\right)$$Can you finish now? we get $$A_{min}=\frac{32}{3\sqrt{3}}$$ for $$a=\frac{2}{\sqrt{3}}$$