A) \[{{l}_{1}},{{l}_{2}},{{l}_{3}}\]are in GGP
B) \[{{l}_{2}},{{l}_{1}},{{l}_{3}}\]are in GP
C) \[{{l}_{3}},{{l}_{1}},{{l}_{2}}\]are in GP
D) \[{{l}_{3}},{{l}_{2}},{{l}_{1}}\]are in GP
Correct Answer: B
Solution :
Let the coordinates of B and C be \[(at_{1}^{2},2a{{t}_{1}})\] and \[(at_{2}^{2},2a{{t}_{2}})\] respectively. Then, the coordinates of A are \[(a{{t}_{1}}\,{{t}_{2}},a({{t}_{1}}+{{t}_{2}})).\] The equation of any tangent to \[{{y}^{2}}=4ax\] is \[ty=x+a{{t}^{2}}.\] \[{{l}_{1}}=\frac{a{{t}_{1}}{{t}_{2}}-a({{t}_{1}}+{{t}_{2}})t+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}},\] \[{{l}_{2}}=\frac{at_{1}^{2}-2a{{t}_{1}}+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}}\] and \[{{l}_{3}}=\frac{at_{2}^{2}-2a{{t}_{2}}+a{{t}^{2}}}{\sqrt{1+{{t}^{2}}}}\] Clearly, \[{{l}_{2}}{{l}_{3}}=l_{1}^{2}\] Therefore, \[{{l}_{2}},{{l}_{1}},{{l}_{3}}\]are in GP.
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