Answer
$\text{Vertex}= (1,1)$
Focus: $f(0,1)$
Directrix is: $x=2$
'Graph b'
Work Step by Step
General form of a horizontal parabola is given as: $(y-k)^2=4p(x-h)$...(1)
Here, $\text{Vertex}=(h,k)$ and focus is: $(h+p, k)$
As we are given $(y-1)^2=-4(x-1)$
From equation (1), we get: $k=1, h=1$ and $4p=-4 \implies p=-1$
$\text{Vertex}=(h,k) \implies (1,1)$
Also, focus is: $(h+p,k) \implies f(1-1,1)=f(0,1)$
Directrix is: $x=h-p \implies x=1+1=2$
Hence, $\text{Vertex}= (1,1)$
Focus: $f(0,1)$
Directrix is: $x=2$
Thus, the 'Graph b' matches with all the above conditions.
