A) \[{{y}^{2}}=a\,(x-a)\]
B) \[{{y}^{2}}=2a\,(x-a)\]
C) \[{{y}^{2}}=4a\,(x-a)\]
D) None of these
Correct Answer: B
Solution :
Any chord PQ which bisected point \[R(h,\,k)\]is \[T=S\]or i.e., \[ky-2a(x+h)={{k}^{2}}-4ah\] Since, it is a focal chord, so it must passes through focus \[(a,\,0)\]. \[\therefore \] \[k(0)-2a(a+h)={{k}^{2}}-4ah\] \[\Rightarrow \] \[{{k}^{2}}=2ah-4{{a}^{2}}\] Hence, locus is \[{{y}^{2}}=2a(x-a)\]
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