Find the vertex, focus, and directrix of the parabola. $( x - 1 ) ^ { 2 } + 16 ( y + 6 ) = 0$
A)Vertex: $( 0 , - 4 )$ ; Focus: $( 1 , - 6 )$ ; Directrix: $y = 0$
B)Vertex: $( 1 , - 10 )$ ; Focus: $( 1 , - 10 )$ ; Directrix: $y = 0$
C)Vertex: $( - 4,1 )$ ; Focus: $( - 4,1 )$ ; Directrix: $y = 0$
D)Vertex: $( 1 , - 6 )$ ; Focus: $( 1 , - 10 )$ ; Directrix: $y = 0$
E)Vertex: $( 1 , - 6 )$ ; Focus: $( 1 , - 4 )$ ; Directrix: $y = 6$

Question

Find the vertex, focus, and directrix of the parabola. $( x - 1 ) ^ { 2 } + 16 ( y + 6 ) = 0$ 
A)Vertex: $( 0 , - 4 )$ ; Focus: $( 1 , - 6 )$ ; Directrix: $y = 0$
B)Vertex: $( 1 , - 10 )$ ; Focus: $( 1 , - 10 )$ ; Directrix: $y = 0$
C)Vertex: $( - 4,1 )$ ; Focus: $( - 4,1 )$ ; Directrix: $y = 0$
D)Vertex: $( 1 , - 6 )$ ; Focus: $( 1 , - 10 )$ ; Directrix: $y = 0$
E)Vertex: $( 1 , - 6 )$ ; Focus: $( 1 , - 4 )$ ; Directrix: $y = 6$

Quizplus · Accepted Answer

The given equation is in the form $(x - h)^2 = 4p(y - k)$, where $(h, k)$ is the vertex. Comparing, we find the vertex is $(1, -6)$. The value of $4p = 16$, so $p = 4$, indicating the focus is 4 units below the vertex since the parabola opens downwards, making the focus $(1, -10)$. The directrix is a horizontal line $p$ units above the vertex, so it is $y = -2$, not $y = 0$ as stated in the options, indicating a mistake in the provided choices.

Find the Vertex, Focus, and Directrix of the Parabola (x−1)2+16(y+6)=0( x - 1 ) ^ { 2 } + 16 ( y + 6 ) = 0(x−1)2+16(y+6)=0

Find the Vertex, Focus, and Directrix of the Parabola $( x - 1 ) ^ { 2 } + 16 ( y + 6 ) = 0$