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=h−kmandOQ=k−hm,
Now,
Area of ΔeaPQ=A=12×OP×OQ =12(h−km)(k−hm)=12[hk−h2m−k2m+hk]=hk−12h2m−k22m⇒dAdm=−12h2+k22m2.∴dAdm=0⇒m2=k2h2⇒m=±kh.Also,d2Adm2=−k2m3>0atm=−kh Hence A is minimum when m =−khand the minimum area =hk+12hk+hl2=2kh
Hence (a) is the correct answer.