Hint: When we get these types of questions, firstly we’ll reduce the given equation to the standard form of that conic and then compare ${x_0}, {y_0}$ and $a$ with the standard equation of parabola. And then find the required parameter by putting values.
Complete step by step answer:
As, we know that the standard equation of parabola is $(y - {y_0})^2 = 4a(x - {x_0})$. In which,
$\Rightarrow$ Vertex = $\left( {{x_0},{y_0}} \right)$ and,
$\Rightarrow$ Focus of parabola is $\left( {{x_0} + a,{y_0}} \right)$
Given Equation of parabola is ${y^2} = 4y - 4x$
First we have to convert the given equation to the standard equation of parabola.
Taking 4y to LHS of the given equation it becomes,
$\Rightarrow {y^2} - 4y = - 4x$
Adding 4 both sides of the equation it becomes,
$\Rightarrow \left( {{y^2} - 4y + 4} \right) = - 4x + 4$
Taking - 4 common in RHS equation becomes,
$\Rightarrow$ $({y - 2})^2 = - 4( x - 1 )$ - (Eq 1)
Comparing equation 1with standard equation of parabola we get,
$\Rightarrow$ ${x_0} = 1, {y_0} = 2$ and $a = - 1$
So, focus of the parabola in equation 1 will be,
$\Rightarrow$ focus = $\left( {1 - 1,2} \right) = \left( {0,2} \right)$
Hence the correct option for the question will be (a).
Note: Understand the diagram properly whenever you are facing these kinds of problems. A better knowledge of formulas will be an added advantage.
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