The triangle of least area through a point

Elimination. From $\frac ap+\frac bq=1$: $\frac bq=\frac{p-a}p$, so $q=\dfrac{bp}{p-a}$. For $q>0$ we need $p>a$.

Area.
$$S(p)=\frac{pq}2=\frac b2\cdot\frac{p^2}{p-a},\qquad p>a.$$

Derivative.
$$S'(p)=\frac b2\cdot\frac{2p(p-a)-p^2}{(p-a)^2}=\frac b2\cdot\frac{p(p-2a)}{(p-a)^2}.$$
It is negative on $(a,2a)$ and positive on $(2a,\infty)$, and $S\to+\infty$ at both ends of $(a,\infty)$: the global minimum is at $p=2a$, hence $q=2b$.

Answer. The minimal area is $S=2ab$, attained when $P$ is the midpoint of the segment of the line between the axes (the midpoint of $(2a,0)$ and $(0,2b)$ is $(a,b)$).

Example. $P=(3,2)$: intercepts $6$ and $4$, area $12$.