Answer
Point $P(x,y)$ with $x=\frac{ab^2}{a^2+b^2}$. $y=\frac{a^2b}{a^2+b^2}$
Work Step by Step
Step 1. Assuming a point $P(x,y)$ on the line, we have $\frac{x}{a}+\frac{y}{b}=1$ or $y=b(1-\frac{x}{a})$
Step 2. The distance from the point to the origin is given by $D=\sqrt {x^2+y^2}$, let $S=D^2=x^2+b^2(1-\frac{x}{a})^2$
Step 3. We need to minimize $D$, which will reach a minimum when $D^2$ is at a minimum.
Step 4. $S'=2x+2b^2(1-\frac{x}{a})(-\frac{1}{a})$; let $S'=0$; we get $a^2x-b^2(a-x)=0$, which gives $x=\frac{ab^2}{a^2+b^2}$. Thus $y=b(1-\frac{x}{a})=b(1-\frac{b^2}{a^2+b^2})=\frac{a^2b}{a^2+b^2}$
Step 5. Check $S''=2-\frac{2b^2}{a}(-\frac{1}{a})=2+\frac{2b^2}{a^2}\gt0$ indicating a concave up region with a minimum distance.
