Find the focus and directrix of the following parabola: \( (y-4)^{2}=16(x-6) \) Focus: \( (10,[?]) \) Directrix: \( x= \)

The equation you provided is in the standard form of a horizontal parabola, \( (y-k)^{2} = 4p(x-h) \), where \( (h, k) \) is the vertex, and \( p \) is the distance from the vertex to the focus (positive in the direction the parabola opens). Here, \( h = 6 \), \( k = 4 \), and \( 4p = 16 \) implies that \( p = 4 \). To find the focus, we start from the vertex \( (h, k) = (6, 4) \). Since the parabola opens to the right, we move \( p \) units to the right. Thus, the focus is at \( (h + p, k) = (6 + 4, 4) = (10, 4) \). The directrix is a vertical line located \( p \) units to the left of the vertex. Therefore, the directrix is \( x = h - p = 6 - 4 = 2 \). Focus: \( (10, 4) \) Directrix: \( x = 2 \)