A) \[4\,(y-2)={{(x-2)}^{2}}\]
B) \[4\,(y+2)={{(x+2)}^{2}}\]
C) \[4\,(y-2)={{(x+2)}^{2}}\]
D) \[(y-2)={{(x+2)}^{2}}\]
Correct Answer: C
Solution :
\[\frac{x-{{t}^{2}}}{1}=\frac{y-2t}{-1}=-2\frac{({{t}^{2}}-2t+2)}{2}\] |
\[\Rightarrow \] \[x=2t-2\] \[\Rightarrow \] \[t=\frac{x+2}{2}\] |
and \[y={{t}^{2}}+2\] \[\Rightarrow \] \[y-2=\frac{{{(x+2)}^{2}}}{4}\] |
\[\Rightarrow \] \[4\,(y-2)={{(x+2)}^{2}}\] |
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