A) (3, 6)
B) \[(9,6\sqrt{3})\]
C) \[(7,2\sqrt{21})\]
D) \[(8,4\sqrt{6})\]
E) \[(1,\sqrt{12})\]
Correct Answer: B
Solution :
We have \[{{y}^{2}}=12x\] ...(i) Comparing Eq. (i) with\[{{y}^{2}}=4ax,\]we get \[4a=12\] \[\Rightarrow \] \[a=3\] \[\therefore \] Focus\[=(3,\text{ }0)\] Let\[({{x}_{1}},{{y}_{1}})\]be point on parabola Eq. (i), hence \[y_{1}^{2}=12{{x}_{1}}\] ...(ii) Now, according to question, \[\sqrt{{{({{x}_{1}}-3)}^{2}}+{{({{y}_{1}}-0)}^{2}}}=12\] \[\Rightarrow \] \[x_{1}^{2}+9-6{{x}_{1}}+y_{1}^{2}=144\] ...(iii) Putting\[y_{1}^{2}=12x,\]from Eq. (ii) in Eq. (iii), we get \[x_{1}^{2}-6{{x}_{1}}+9+12{{x}_{1}}=144\] \[\Rightarrow \] \[x_{1}^{2}+6{{x}_{1}}-135=0\] \[\Rightarrow \] \[({{x}_{1}}+15)({{x}_{1}}-9)=0\] \[\Rightarrow \] \[{{a}_{1}}=9,-15\] \[{{x}_{1}}=-15\]is not possible. From Eq. (i) \[y_{1}^{2}=12{{x}_{1}}\] \[\Rightarrow \] \[y_{1}^{2}=12\times 9=108\] \[\Rightarrow \] \[{{y}_{1}}=6\sqrt{3}\] \[\therefore \]Required point\[=(9,6\sqrt{3})\].
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