Correct Answer - `3y^(2)=16ax`
Let the point be (h,k). Let any tangent be
`y=mx+(a)/(m)`
`or" "k=mh+(a)/(m)orm^(2)h-mk+a=0`
Its roots are `m^(1)and3m_(1)`. Therefore,
`m_(1)+3m_(1)=(k)/(h)`
`m_(1)*3m_(1)=(a)/(h)`
Eliminating `m^(1)`, we get the locus as
`3y^(2)=16ax`