question_answer

If the line \[\frac{x}{a}+\frac{y}{b}=1\]  moves such that \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}},\] where c is a constant, then the locus of the foot of the perpendicular from the origin to the line is:

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.