A) \[3\]
B) \[4\]
C) \[9/2\]
D) \[6\]
Correct Answer: B
Solution :
[b] Let the coordinates of point P be (h, k). Equation of any normal to parabola \[{{y}^{2}}=4x\]is \[y=mx-2m-{{m}^{3}}\]. It passes through \[P(h,k)\]. \[\Rightarrow \,\,\,\,\,\,\,\,{{m}^{3}}+m(2-h)+k=0,\] which has roots \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\]. \[\Rightarrow \,\,\,\,\,\,\,\,\,{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0,\] \[{{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}=2-h,\] and \[{{m}_{1}}{{m}_{2}}{{m}_{3}}=-k\] Let the points on the parabola be \[({{x}_{1}},{{y}_{1}})\equiv ({{m}_{1}}^{2},-2{{m}_{1}}),\] \[({{x}_{2}},{{y}_{2}})\equiv ({{m}_{2}}^{2},-2{{m}_{2}}),\] \[({{x}_{3}},{{y}_{3}})\equiv ({{m}_{3}}^{2},-2{{m}_{3}}),\] Given that \[{{y}_{1}}{{y}_{2}}+{{y}_{2}}{{y}_{3}}+{{y}_{3}}{{y}_{1}}={{x}_{1}}{{x}_{2}}{{x}_{3}}\] \[\therefore \,\,4({{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}})={{m}_{1}}^{2}.{{m}_{2}}^{2}.{{m}_{3}}^{2}\] \[\therefore \,\,\,\,\,4(2-h)={{k}^{2}}\] \[\therefore \] Locus of \[(h,k)\] is \[{{y}^{2}}=-4(x-2),\] which is a parabola having length of latus rectum 4.
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