The point of the curve \[{{y}^{2}}=2(x-3)\] at which the normal is parallel to the line\[y-2x+1=0\]is [MP PET 1998]

\[\left( \frac{3}{2},\,2 \right)\]

\[\left( -\frac{1}{2},-2 \right)\]

The point of the curve \[{{y}^{2}}=2(x-3)\] at which the normal is par

1.	The point of the curve \[{{y}^{2}}=2(x-3)\] at which the normal is parallel to the line\[y-2x+1=0\]is [MP PET 1998]
A.	(5,2)
B.	\[\left( -\frac{1}{2},-2 \right)\]
C.	(5, ?2)
D.	\[\left( \frac{3}{2},\,2 \right)\]
Answer» D. \[\left( \frac{3}{2},\,2 \right)\]

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The point of the curve \[{{y}^{2}}=2(x-3)\] at which the normal is parallel to the line\[y-2x+1=0\]is [MP PET 1998]

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