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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 Δ OPQ, O being the origin, is


A

2hk

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B

kh

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C

4kh

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D

3hk

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Solution

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.

OP=hkmand OQ=khm,
Now,
Area of Δ ea PQ=A=12×OP×OQ
=12(hkm)(khm)=12[hkh2mk2m+hk]=hk12h2 mk22mdAdm=12h2+k22m2.dAdm=0m2=k2h2m=±kh.Also,d2Adm2=k2m3>0 at m=kh
Hence A is minimum when m =kh and the minimum area
=hk+12hk+hl2=2kh

Hence (a) is the correct answer.

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