Question

Let a variable line passing through a fixed point $P$ in the first quadrant cuts the positive coordinate axes at points $A$ and $B$ respectively. If the area of $\Delta OAB$ is minimum, then $OP$ is

• A. Altitude through vertex $O$ of $\Delta AOB$
• B. Median through vertex $O$ of $\Delta AOB$
• C. Internal angle bisector through vertex $O$ of $\Delta AOB$
• D. None of these

Answer 1

Answer: (B)

Let, the foot of perpendiculars from $P$ on $OA$ & $OB$ are $C$ & $D$ respectively
Let, $\angle PAC=\theta =\angle BPD$

Now, $CA=kcot\theta$ $\&$ $BD=htan\theta$
Area of $\Delta OAB=$ area of $\Delta PCA+$ area of $\Delta BPD+$ area of the rectangle $PDOC$
$\Rightarrow$ Area of $\Delta OAB=\frac{1}{2}{k}^{2}cot\theta +\frac{1}{2}{h}^{2}tan\theta +hk=\frac{1}{2}{\left(k\sqrt{cot\theta }-h\sqrt{tan\theta }\right)}^2+2hk$
$\Rightarrow$ The minimum area of $\Delta OAB$ is $2hk$ when $k\sqrt{cot\theta }=h\sqrt{tan\theta }$ $\Rightarrow tan\theta =\frac{k}{h}$
$\Rightarrow hsin\theta =kcos\theta \Rightarrow hsec\theta =kcosec\theta$
$\Rightarrow PB=PA\Rightarrow OP$ is the median