A) \[{{x}^{2}}=4y\]
B) \[{{x}^{2}}=2y\]
C) \[{{x}^{2}}=16y\]
D) \[{{y}^{2}}=2x\]
Correct Answer: D
Solution :
Since, O be the origin and A be the point on the curve\[{{y}^{2}}=4x\]. \[\therefore \]Coordinates of O and A are (0, 0) and \[(a{{t}^{2}},2at)\] respectively. \[\therefore \]Coordinates of mid point of OA are \[\left( \frac{0+a{{t}^{2}}}{2},\frac{0+2at}{2} \right)=\left( \frac{a{{t}^{2}}}{2},at \right)\] Since,\[{{(at)}^{2}}=4\left( \frac{a{{t}^{2}}}{2} \right)\] Hence, the locus of required point is \[{{y}^{2}}=2x.\]
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