A) \[{{y}_{1}},\ {{y}_{2}},\ {{y}_{3}}\] are in A.P.
B) \[{{y}_{1}},\ {{y}_{3}},\ {{y}_{2}}\] are in A.P.
C) \[{{y}_{1}},\ {{y}_{2}},\ {{y}_{3}}\] are in G.P.
D) \[{{y}_{1}},\ {{y}_{3}},\ {{y}_{2}}\] are in G.P.
Correct Answer: B
Solution :
Let the co-ordinates of P and Q be \[(at_{1}^{2},\,2a{{t}_{1}})\]and \[(at_{2}^{2},\,2a{{t}_{2}})\]respectively. Then \[{{y}_{1}}=2a{{t}_{1}}\]and \[{{y}_{2}}=2a{{t}_{2}}.\] The co-ordinates of the point of intersection of the tangents at \[P\]and Q are \[\{a{{t}_{1}}{{t}_{2}},a({{t}_{1}}+{{t}_{2}})\}\] \[\therefore {{y}_{3}}=a({{t}_{1}}+{{t}_{2}})\] Þ \[{{y}_{3}}=\frac{{{y}_{1}}+{{y}_{2}}}{2}\]Þ \[{{y}_{1}},\,{{y}_{3}}\] and \[{{y}_{2}}\]are in A.P.
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