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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