Let a line passing through the fixed point $(10, 5)$ cuts the $x$-axis at $P$ and $y$-axis at $Q$. T

Question

Let a line passing through the fixed point $(10, 5)$ cuts the $x$-axis at $P$ and $y$-axis at $Q$. Then find the minimum area of $	riangle O P Q$.

Professor PC Thomas Classes · Answer

Let $P(a, 0), Q(0, b)$ Equation of $P Q$ is $\frac{x}{a}+\frac{y}{b}=1$\ This passes through $(h, k)$\ $	herefore \frac{h}{a}+\frac{k}{b}=1 \quad \Rightarrow b=\frac{a k}{a-h}$\ Area of $	riangle O P Q=\frac{1}{2} a b=\frac{1}{2} a\left(\frac{a k}{a-h}ight)=s$ (say)\ For $s$ to be minimum,\ $\frac{d s}{d a}=\frac{1}{2} k\left[\frac{(a-h) 2 a-a^{2}}{(a-h)^{2}}ight]=0$\ $\Rightarrow \quad a=2 h$\ Now $\frac{d^{2} s}{d a^{2}}=\frac{2}{h}$ for $a=2 h$, which is positive Hence $s$ is minimum when $a=2 h$ and Minimum area of $	riangle O P Q=\frac{k}{2}\left[\frac{4 h^{2}}{h}ight]=2 k h$\