11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer The locus of mid point of that chord of parabola which subtends right angle on the vertex will be   [UPSEAT 1999]

    A)            \[{{y}^{2}}-2ax+8{{a}^{2}}=0\] 

    B)            \[{{y}^{2}}=a(x-4a)\]

    C)            \[{{y}^{2}}=4a(x-4a)\]               

    D)            \[{{y}^{2}}+3ax+4{{a}^{2}}=0\]

    Correct Answer: A

    Solution :

               Equation of parabola \[{{y}^{2}}=4ax\]                       ?..(i)            Equation of that chord of parabola whose mid point is \[({{x}_{1}},\,{{y}_{1}})\] will be \[y{{y}_{1}}-2a(x+{{x}_{1}})=y_{1}^{2}-4a{{x}_{1}}\]            or \[y{{y}_{1}}-2ax=y_{1}^{2}-2a{{x}_{1}}\] or \[\frac{y{{y}_{1}}-2ax}{y_{1}^{2}-2a{{x}_{1}}}=1\]    .....(ii)            Making equation (i) homogeneous by equation (ii), the equation of lines joining the vertex (0,0) of parabola to the point of intersection of chord (ii) and parabola (i) will be            \[{{y}^{2}}=4ax\frac{y{{y}_{1}}-2ax}{y_{1}^{2}-2a{{x}_{1}}}\] or \[{{y}^{2}}(y_{1}^{2}-2a{{x}_{1}})=4ax(y{{y}_{1}}-2ax)\]            or \[8{{a}^{2}}{{x}^{2}}-4a{{y}_{1}}xy+(y_{1}^{2}-2a{{x}_{1}}){{y}^{2}}=0\]            If lines represented by it are mutually perpendicular, then coefficient of \[{{x}^{2}}+\] coefficient of \[{{y}^{2}}=0\]            therefore, \[8{{a}^{2}}+(y_{1}^{2}-2a{{x}_{1}})=0\] or \[y_{1}^{2}-2a{{x}_{1}}+8{{a}^{2}}=0\].                    \ Required locus of \[({{x}_{1}},\,{{y}_{1}})\] is \[{{y}^{2}}-2ax+8{{a}^{2}}=0.\]


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