A) \[{{\operatorname{x}}^{2}}+{{y}^{2}}={{c}^{2}}/2\]
B) \[{{\operatorname{x}}^{2}}+{{y}^{2}}=2{{c}^{2}}\]
C) \[{{\operatorname{x}}^{2}}+{{y}^{2}}={{c}^{2}}\]
D) \[{{\operatorname{x}}^{2}}-{{y}^{2}}={{c}^{2}}\]
Correct Answer: C
Solution :
Equation of perpendicular drawn from origin to the line \[\frac{x}{a}+\frac{y}{b}=1\] is, \[y-0=\frac{a}{b}(x-0)\] \[\Rightarrow \,\,\,\operatorname{by}-ax=0\,\,\Rightarrow \,\,\frac{x}{b}-\frac{y}{a}=0\] The locus of foot of perpendicular is the intersection point of line \[\frac{x}{a}+\frac{y}{b}=1\] ... (i) and \[\frac{x}{b}-\frac{y}{a}=0\] ... (ii) Squaring and adding (i) and (ii), we get \[{{\left( \frac{x}{a}+\frac{y}{b} \right)}^{2}}+{{\left( \frac{x}{b}-\frac{y}{a} \right)}^{2}}=1\] \[\Rightarrow \,\,\,\,{{x}^{2}}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)+\,\,{{y}^{2}}{{\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)}^{2}}=1\] \[=\,\,\,{{x}^{2}}\left( \frac{1}{{{c}^{2}}} \right)+{{y}^{2}}\left( \frac{1}{{{c}^{2}}} \right)=1\,\,\,\left[ \because \,\,\,\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}} \right]\] \[\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]
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