Find a smooth curve in the plane passing through the point $(1,0)$ such that for every point $(a,b)$ on the curve, the tangent line to the curve at $(a,b)$ intersects the $y$-axis at the point $(0,a)$.
The tangent line through the point $(1,0)$ is $y=-x+1$. In general, the tangent line to the curve at point $(a,b)$ is $y-b=f'(a)(x-a)$.
So, when $x=0$, the tangent line is $y=b-a\cdot f'(a)$. What we want is for when $x=0$, $y=a$. So, we would have $a=b-af'(a)$. So, we want $f'(a)=\frac{b-a}{a}$.
How can I proceed?