A) \[{{y}^{2}}-2y-12x-11=0\]
B) \[{{x}^{2}}+2x-12y+13=0\]
C) \[{{y}^{2}}-2y+12x+11=0\]
D) \[{{y}^{2}}-2y-12x+13=0\]
Correct Answer: A
Solution :
By the condition of parabola \[P{{M}^{2}}=P{{S}^{2}}\] \[\Rightarrow \] \[{{(x+4)}^{4}}={{(x-2)}^{2}}+{{(y-1)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+8x+16={{x}^{2}}-4x+4+{{y}^{2}}-2y+1\] \[\Rightarrow \] \[{{y}^{2}}-2y-12x-11=0\]You need to login to perform this action.
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