question_answer

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. Then the minimum area of the \[\Delta OPQ.O\] O being the origin, is

A) 4hk sq. units

B) 2hk sq. units

C) 3hk sq. units

D) None of these

Correct Answer: B

Solution :

[b] Let the equation of any line passing through \[A(h,k)\] be \[y-k=m(x-h).\]

Let this line cut the x-axis and y-axis at P and Q.

Then \[P\equiv \left( h-\frac{k}{m},0 \right)\] and \[Q\equiv (0,k-mh).\]

Let S be the area of \[\Delta OPQ,\] then

\[S=\frac{1}{2}OP\times OQ=\frac{1}{2}\left( h-\frac{k}{m} \right)(k-mh)\]

\[=\frac{1}{2}\frac{(mh-k)(k-mh)}{m}\]

\[\Rightarrow 2mS=hkm-{{k}^{2}}-{{h}^{2}}{{m}^{2}}+khm\]

\[\Rightarrow {{h}^{2}}{{m}^{2}}-2(hk-S)m+{{k}^{2}}=0\]

Since, m is real \[\therefore \] its discriminant \[D\ge 0\]

\[\therefore 4{{(hk-S)}^{2}}-4{{h}^{2}}{{k}^{2}}\ge 0\]

\[\Rightarrow S-2hk\ge 0\Rightarrow S\ge 2hk\]

Hence, minimum value of S is 2hk sq. units.