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 
0AB is minimum, then OP is
AOB
AOB
AOB
Text Solution
Verified by ExpertsB
Median through vertex O of 
AOB
Let, the foot of perpendiculars from P on OA &OB are C & D respectively
Let, 
PAG = 
= 
BPD
Now, CA = k cot 
&BD = h tan 
Area of 
OAB = area of 
PCM+ area of 
BPD+ area of the rectangle PDOC
Area of 
OAB = 
k 2 cot 
+ 
h 2 tan 
+ hk = 
(k 
- h 
) 2 + 2hk
The minimum area of 
OAB is 2hk when 
h sin 
= k cos 
h sec 
= k cosec 
PB = PA 
OP is the median
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