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The vertex of the parabola (y-2)^2=16(x-...

The vertex of the parabola `(y-2)^2=16(x-1)` is a.`(1,2)` b.`(-1,2)` c. `(1,-2)` d. `(2,1)`

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given parabola may be written as `sqrt((x-1)^2 +(y-2)^2) = |(x+y+3)|/sqrt3`
focus is`(1,2)` directrix is `x+y+3=0`
thus the equation of the axis is ` x-y+1=0`
solving both above we get ` p(-2,-1)` ...
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