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: C
Solution :
\[P=(1,\,0),Q=(h,k)\] such that \[{{k}^{2}}=8h\] Let \[(\alpha ,\beta )\] be the midpoint of \[PQ\]; \[\alpha =\frac{h+1}{2},\beta =\frac{k+0}{2};\,\,2\alpha -1=h,\,\,2\beta =k\] \[{{(2\beta )}^{2}}=8(2\alpha -1)\Rightarrow {{\beta }^{2}}=4\alpha -2\Rightarrow {{y}^{2}}-4x+2=0\].
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