Question

Let $P(h,k)$ be a fixed point where $h,k>0$. A straight line passing though this points $P(h,k)$ cut the positive direction of co-ordinates axes at the points $Q$ and $R$. Find the minimum are of the $\Delta OQR$,$O$ being the $(0,0)$

Answer 1

Let $Q$ be $(a,0)$ and $R(0,b)$

$\therefore$ equation of the intercept $QR\Rightarrow \frac{x}{a}+\frac{y}{b}=1$

$\therefore \frac{h}{a}+\frac{k}{b}=1(\because (h,k)$ is $a$ point on the line)

$\Rightarrow b=\frac{ak}{a-h}$

$\therefore$ Area of $△OQR=\frac{1}{2}ab=\frac{1}{2}a.\frac{ak}{a-h}=\frac{1}{2}\frac{k}{\frac{a-h}{a^2}}$

$\therefore$ Area is min, when $y=\frac{a-h}{a^2}=\frac{1}{a}-\frac{h}{a^2}$ is max

$\therefore \frac{dy}{da}=\frac{-1}{a^2}+\frac{2h}{a^3}=0\Rightarrow a=2h$

$\therefore$ minimum area of $△OQR=\frac{1}{2}\frac{k.h^2}{2h-h}=\frac{1}{2}hk$