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