Question

Let (h, k) be a fixed point, where h > 0, k > 0. A straight line passing through this point cuts the positive direction of the coordinate axes at the points P and Q. The minimum area of the $\Delta$ OPQ, O being the origin, is

• A. 2hk
• B. kh
• C. 4kh
• D. 3hk

Answer 1

The correct option is A

2hk

Let the equation of the line through (h, k) be y - k = m (x - h), where m is the slope of the line.

$\therefore OP=h-\frac{k}{m}andOQ=k-hm,$
Now,
Area of $\Delta eaPQ=A=\frac{1}{2}\times OP\times OQ$
$=\frac{1}{2}(h-\frac{k}{m})(k-hm)=\frac{1}{2}[hk-h^{2}m-\frac{k^2}{m}+hk]=hk-\frac{1}{2}{h}^{2}m-\frac{k^2}{2m}\Rightarrow \frac{dA}{dm}=-\frac{1}{2}{h}^2+\frac{k^2}{2m^2}.\therefore \frac{dA}{dm}=0\Rightarrow m^2=\frac{k^2}{h^2}\Rightarrow m=\pm \frac{k}{h}.Also,\frac{d^{2}A}{d{m}^2}=-\frac{k^2}{m^3}>0atm=-\frac{k}{h}$
Hence A is minimum when m $=-\frac{k}{h}$ and the minimum area
$=hk+\frac{1}{2}hk+\frac{hl}{2}=2kh$
Hence (a) is the correct answer.