Question

A straight line passing through the point $(87,33)$ cuts the positive direction of the coordinate axes at the points $P$ and $Q$. If $O$ is the origin and the minimum area of the triangle $OPQ$ is $k$, then find the value of $\frac{k}{826}$.

Answer 1

Equation of a line passing through $(87,33)$ is $y-33=m(x-87)$ cuts y-axis at $P(0,33-87m)$ and x-axis at $(87-\frac{33}{m},0)$
Area of triangle OPQ,A $=\left|\frac{1}{2m}\right(33-87m\left)\right(87m-33\left)\right|$
$\Rightarrow A=-\frac{1}{2}m(87{)}^2+33\times 87-\frac{1}{2m}(33{)}^2$
$\frac{dA}{dm}=-\frac{1}{2}(87{)}^2+\frac{1}{2m^2}(33{)}^2=0\Rightarrow m^2=\frac{(33{)}^2}{(87{)}^2}$
$m=\pm \frac{33}{87}$
$\frac{d^{2}A}{d{m}^2}=-\frac{1}{m^3}(33{)}^2>0ifm=-\frac{33}{87}$

So, the minimum value of $A$ is when $m=-\frac{33}{87}$ and the required value is
$k=33\times 87+\frac{1}{2}[\frac{33}{87}\times {87}^2+\frac{87}{33}\times {33}^2]$
$k=2\times 33\times 87=5742$

Now, $\frac{k}{826}=\frac{5742}{826}=7$