A) \[{{a}^{2}}+{{b}^{2}}\]
B) \[{{a}^{2}}-{{b}^{2}}\]
C) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]
D) \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]
Correct Answer: C
Solution :
The given points are \[\left( a\,\cos \theta +b\,\sin \theta ,\,0 \right)\] and \[\left( 0,\,a\,\sin \theta -b\,\cos \theta \right)\]. |
Here, \[{{x}_{1}}=a\,\cos \theta +b\,\sin \theta ,\,{{y}_{1}}=0\] |
\[{{x}_{2}}=0,\,{{y}_{2}}=a\,\sin \theta -b\,\cos \theta \] |
\[\therefore \] Required distance |
\[=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] |
\[=\sqrt{{{\left( a\,\cos \,\theta +b\,\sin \theta \right)}^{2}}+{{\left( b\,\cos \theta -a\,\sin \theta \right)}^{2}}}\] |
\[=\sqrt{{{\cos }^{2}}\theta \left( {{a}^{2}}+{{b}^{2}} \right)+{{\sin }^{2}}\theta \left( {{a}^{2}}+{{b}^{2}} \right)}\] |
\[=\sqrt{{{a}^{2}}+{{b}^{2}}}\] |
\[\left[ \because \,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \right]\] |
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