A) \[(4,4)\]
B) \[(8,16)\]
C) \[(1/4,1)\]
D) None of these
Correct Answer: C
Solution :
[c] \[(4,-4)\] and \[(9,6)\] lie on \[{{y}^{2}}-4x(x-b)\] \[\Rightarrow \,\,\,\,\,16=4a(4-b)\] and \[36=4a(9-b)\] Solving these, we get \[a=1,\text{ }b=0\] So, equation of parabola is \[{{y}^{2}}=4x.\] Let the point R be \[({{t}^{2}},2t)\], where \[t\in (-2,3)\]. \[=\frac{1}{2}\left| 10t-10{{t}^{2}}+60 \right|\,\,=\frac{1}{4}\,\,\left| 125-5{{(2t-1)}^{2}} \right|\] Thus, area is largest when \[t=\frac{1}{2}\] \[\therefore \,\,\,\,\,\,\,\,R({{t}^{2}},2t)=R(1/4,1)\]You need to login to perform this action.
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