JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer The equation of the locus of foot of perpendiculars drawn from the origin to the line passing through a fixed point     (a, b), is

    A)            \[{{x}^{2}}+{{y}^{2}}-ax-by=0\]                                       

    B)            \[{{x}^{2}}+{{y}^{2}}+ax+by=0\]

    C)            \[{{x}^{2}}+{{y}^{2}}-2ax-2by=0\]                                  

    D)            None of these

    Correct Answer: A

    Solution :

               \[\lambda (x-a)+(y-b)=0\]is the equation of line.                    \[r=-\left( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} \right)\]            Coordinates of point \[\equiv \left\{ -\lambda \left( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} \right)\text{ },-\left( \frac{-a\lambda -b}{{{\lambda }^{2}}+1} \right) \right\}\]            \[h=\lambda \left( \frac{a\lambda +b}{{{\lambda }^{2}}+1} \right)\text{ },k=\frac{a\lambda +b}{{{\lambda }^{2}}+1},\lambda =\frac{h}{k}\]            \[\therefore h=h\left( \frac{ah+kb}{{{h}^{2}}+{{k}^{2}}} \right)\Rightarrow {{x}^{2}}+{{y}^{2}}=ax+by.\]


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