A) \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}/2\]
B) \[{{x}^{2}}+{{y}^{2}}=2{{c}^{2}}\]
C) \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]
D) \[{{x}^{2}}-{{y}^{2}}={{c}^{2}}\]
Correct Answer: C
Solution :
Equation of perpendicular drawn from origin to the line \[\frac{x}{a}+\frac{y}{b}=1\] is \[y-0=\frac{a}{b}(x-0)\] \[\left[ \begin{align} & \\ & \because m \\ \end{align} \right.\] of given line \[=\frac{-b}{a}\], \[\therefore m\] of perpendicular \[\left. =\frac{a}{b} \right]\] Þ \[by-ax=0\] Þ \[\frac{x}{b}-\frac{y}{a}=0\] Now, 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) To find locus, squaring and adding (i) and (ii) \[{{\left( \frac{x}{a}+\frac{y}{b} \right)}^{2}}+{{\left( \frac{x}{b}-\frac{y}{a} \right)}^{2}}=1\] Þ \[{{x}^{2}}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)+{{y}^{2}}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)=1\] Þ \[{{x}^{2}}\left( \frac{1}{{{c}^{2}}} \right)+{{y}^{2}}\left( \frac{1}{{{c}^{2}}} \right)=1\] , \[\left[ \because \,\,\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}} \right]\] Þ \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\].
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