A) straight line
B) circle
C) parabola
D) ellipse
Correct Answer: B
Solution :
Equation of line is \[\frac{x}{a}+\frac{y}{b}=1\] Let P be the foot of perpendicular from the origin to the line whose coordinate is \[({{x}_{1}},{{y}_{1}}).\] Since \[OP\bot AB\] \[\therefore \] slope of \[OP\times \]slope of \[AB=-1\] \[\Rightarrow \] \[\left( \frac{{{y}_{1}}}{{{x}_{1}}} \right)\left( \frac{b}{-a} \right)=-1\] \[\Rightarrow \] \[b{{y}_{1}}=a{{x}_{1}}\] ?(i) Since P lies on the line AB, then \[\frac{{{x}_{1}}}{a}+\frac{{{y}_{1}}}{b}=1\] \[\Rightarrow \] \[b{{x}_{1}}+a{{y}_{1}}=ab\] ?(ii) From Eqs. (i) and (ii), \[{{x}_{1}}=\frac{a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]and \[{{y}_{1}}=\frac{{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}}\] Now, \[x_{1}^{2}+y_{1}^{2}={{\left( \frac{a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \frac{{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}\] \[\Rightarrow \] \[x_{1}^{2}+y_{1}^{2}=\frac{{{a}^{2}}{{b}^{2}}}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}+\frac{{{a}^{4}}{{b}^{2}}}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}\] \[\Rightarrow \] \[x_{1}^{2}+y_{1}^{2}=\frac{{{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}\] \[\Rightarrow \] \[x_{1}^{2}+y_{1}^{2}=\frac{{{a}^{2}}{{b}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] \[\Rightarrow \] \[x_{1}^{2}+y_{1}^{2}=\frac{1}{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}\] But \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\] (given) \[\therefore \] \[x_{1}^{2}+{{y}_{1}}^{2}={{c}^{2}}\] Thus, the locus of \[P({{x}_{1}},{{y}_{1}})\]is \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\] which is the equation of circle.You need to login to perform this action.
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