A) \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]
B) \[{{x}^{2}}+{{y}^{2}}=2{{c}^{2}}\]
C) \[{{x}^{2}}+{{y}^{2}}=\frac{c}{2}\]
D) None of these
Correct Answer: A
Solution :
Given equation of line is \[\frac{x}{a}+\frac{y}{b}=1\] \[\Rightarrow \] \[bx+ay=ab\] where \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\] Equation of perpendicular from origin is \[ax-by=0\] Let the coordinates of the foot of the perpendicular line be\[(h,k)\]. \[\therefore \] \[bh+ak=ab\] ...(i) and \[ah-bk=0\] ?..(ii) On solving Eqs. (i) and (ii), we get \[h=\frac{a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] and \[k=\frac{{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}}\] \[\therefore \] \[{{h}^{2}}+{{k}^{2}}={{\left( \frac{a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \frac{{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}\] \[=\frac{{{a}^{2}}{{b}^{4}}+{{a}^{4}}{{b}^{2}}}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}\] \[=\frac{{{a}^{2}}{{b}^{2}}({{b}^{2}}+{{a}^{2}})}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}=\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[=\frac{1}{\frac{1}{{{b}^{2}}}+\frac{1}{{{a}^{2}}}}=\frac{1}{\frac{1}{{{c}^{2}}}}\] \[\Rightarrow \] \[{{h}^{2}}+{{k}^{2}}={{c}^{2}}\] Hence, required locus is \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]
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