A) A.P
B) GP.
C) H.P.
D) None of these
Correct Answer: B
Solution :
[b] Let \[P(a{{t}^{2}}_{1},2a{{t}_{1}})\] and \[Q(a{{t}^{2}}_{2},2a{{t}_{2}})\] to two points on the parabola \[{{y}^{2}}=4ax.\]The tangents at P and Q intersect at \[T[a{{t}_{1}}{{t}_{2}},a({{t}_{1}}+{{t}_{2}})].\] Now, \[SP=\sqrt{{{(at_{1}^{2}\,-\,\,a)}^{2}}+2{{(a{{t}_{1}}-0)}^{2}}}\] \[=a({{t}^{2}}_{1}+1);\] \[SQ=a({{t}^{2}}_{2}+1)\] and \[ST=\sqrt{{{(a{{t}_{1}}{{t}_{2}}-a)}^{2}}+{{[a({{t}_{1}}+{{t}_{2}})-0]}^{2}}}\] \[=a\sqrt{(1+{{t}^{2}}_{1})(1+{{t}^{2}}_{2})}\] \[\Rightarrow S{{T}^{2}}={{a}^{2}}(1+{{t}^{2}}_{1})(1+{{t}^{2}}_{2})=SP.SQ\] Hence SP.ST, SQ are in G.P.
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