Determine a curve such that, at every point, the tangent exists and is not horizontal. The segment of this tangent between the axis has as middle point, the contact point of the curve, and the curve touches the points $$(2,4), (-2,4)$$
I'm having trouble with this problem. I have tried to use the midpoint theorem, but I don't see it.
Tangent $AB$ has equation $$Y-y=y'(X-x)$$ we have axis intersection points $A\left(\frac{xy'-y}{y'},0\right)$ and $B\left(0,y-xy'\right)$
so the midpoint is $$P\left(\frac{xy'-y}{2y'},\frac{y-xy'}{2}\right)$$ as $P$ is on the curve we must have $$\begin{cases} x=\frac{xy'-y}{2y'}\to y+xy'=0\\ y=\frac{y-xy'}{2}\\ \end{cases} $$ Consider the second equation, which is equivalent to the first $$2y=y-xy'\to \frac{y'}{y}=-\frac{1}{x}\to \log y =-\log x+C\to y=\frac{k}{x}$$ as the curve must pass through $(2,4)$ we have $k=8$ so the solution is $$y=\frac{8}{x}$$