A) \[{{y}^{2}}=3x\]
B) \[{{y}^{2}}=2x\]
C) \[{{y}^{2}}=12x\]
D) \[{{y}^{2}}=6x\]
Correct Answer: C
Solution :
\[\because \]\[S{{P}^{2}}=P{{M}^{2}}\] Þ\[{{(x-3)}^{2}}+{{y}^{2}}={{\left| \frac{x+3}{\sqrt{1}} \right|}^{2}}\] Þ\[{{x}^{2}}+9-6x+{{y}^{2}}={{x}^{2}}+9+6x\] Þ\[{{y}^{2}}=12x\].You need to login to perform this action.
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