Identify the focus and directrix of the parabola whose equation is \( (y-3)^{2}=-4(x-5) \). (1 point) focus: \( (4,2) \); directrix: \( y=4 \) focus: \( (4,4) \); directrix: \( y=2 \) focus: \( (4,3) \); directrix: \( x=6 \) focus: \( (6,3) \); directrix: \( x=4 \)

To identify the focus and directrix of the given parabola, we start with the equation \( (y-3)^{2} = -4(x-5) \). This can be rewritten in standard form, indicating that it opens to the left since the coefficient of \((x-5)\) is negative. The vertex of the parabola is at the point \((5, 3)\). For parabolas in the form \((y-k)^2 = 4p(x-h)\), we see that \(4p = -4\), which means \(p = -1\). The focus, located \(p\) units from the vertex, is then at \((5 + p, 3) = (5 - 1, 3) = (4, 3)\), and the directrix, being a vertical line \(p\) units to the right of the vertex, is given by \(x = 5 + 1\) or \(x = 6\). So, the focus is at \((4, 3)\) and the directrix is \(x = 6\). Thus, the correct choice is: focus: \( (4, 3) \); directrix: \( x=6 \).