A) \[{{x}^{2}}-4y+2=0\]
B) \[{{x}^{2}}+4y+2=0\]
C) \[{{y}^{2}}+4x+2=0\]
D) \[{{y}^{2}}-4x+2=0\]
Correct Answer: D
Solution :
The coordinates of P are (1, 0). A general point Q on \[{{y}^{2}}=8x\] is\[(2{{t}^{2}},4t)\]. Mid-point of PO is (h, k) so \[2h=2{{t}^{2}}+1\] ?.(i) and \[2k=4t\Rightarrow t=k/2\] ...(ii) On putting the value of t from Eq. (ii) in Eq, (i), we get \[2h=\frac{2{{k}^{2}}}{4}+1\Rightarrow 4h={{k}^{2}}+2\] Hence, the locus of (h, k) is\[{{y}^{2}}-4x+2=0.\]
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