A) \[4{{y}^{2}}=25x\]
B) \[2{{y}^{2}}=25x\]
C) \[3{{y}^{2}}=16x\]
D) \[3{{y}^{2}}=20\]
Correct Answer: A
Solution :
[a] Let \[P(a{{t}_{1}}^{2},2a{{t}_{1}}),\,\,Q(a{{t}_{2}}^{2},2a{{t}_{2}})\] and \[T(h,k)\] Here, \[h={{t}_{1}}{{t}_{2}},\,\,\,k={{t}_{1}}+{{t}_{2}}.\] Also, \[{{t}_{1}}^{2}=16{{t}_{2}}^{2}\] (given) Eliminating \[{{t}_{1}}\] and \[{{t}_{2}},\] we get locus of \[T(h,k)\] as \[4{{y}^{2}}=25x.\]You need to login to perform this action.
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