Equation of the parabola with its vertex at \[\left( {1,1} \right)\] and focus \[\left( {3,1} \right)\] is
A. ${(x - 1)^2} = 8(y - 1)$
B. ${(y - 1)^2} = 8(x - 3)$
C. ${(y - 1)^2} = 8(x - 1)$
D. ${(x - 3)^2} = 8(y - 1)$

Here,
 \[V \to \] vertex of the parabola
 \[F \to \] focus of the parabola

And as the vertex is equidistant from foci and directrix, and latter is perpendicular to the axis of symmetry.

We get the equation of directrix as
 \[ \Rightarrow X = - 1\]
Hence, the directrix is \[X + 1 = 0\]
We know that equation of a parabola whose focus \[F\left( {a,b} \right)\] and directrix \[X + 1 = 0\] is know is given by
 \[{(x - a)^2} + {(y - b)^2} = directri{x^2}\] … (1)

Parabola is the locus of a point whose distance from directrix \[x + 1 = 0\] and focus \[(3,1)\] , Hence
We get \[a = 3\] , \[b = 1\] and directrix \[x = X + 1\] put values in (1), we get
 \[{(x - 3)^2} + {(y - 1)^2} = {(x + 1)^2}\] … (2)

We know that,
 \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
 \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]

Hence, further solving (2), we get
 \[ \Rightarrow {x^2} + 9 - 6x + {y^2} + 1 - 2y = {x^2} + 1 + 2x\]

Cancelling out equal terms from, LHS and RHS, we get
 \[ \Rightarrow 9 - 8x + {y^2} - 2y = 0\]
On rearranging we get,
 \[ \Rightarrow {y^2} - 2y + 9 = 8x\]

Completing the square in LHS, we get
 \[ \Rightarrow {y^2} - 2y + 1 + 8 = 8x\]
Using \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] , we get,
 \[ \Rightarrow {(y - 1)^2} = 8x - 8\]
Taking 8 common on RHS we get,
 \[ \Rightarrow {(y - 1)^2} = 8(x - 1)\]

Hence, the required equation is \[{(y - 1)^2} = 8(x - 1)\]
Therefore, the final answer is option C.

Note:
These types of questions can be easily solved if we can visually illustrate the required curve in a 2D plane because after that many hidden information is gained, for example in this particular question, the equation of directrix was found as a key to solve the given question easily.