Find the vertex and focus of the parabola. $x ^ { 2 } + 12 y = 0$
A)vertex: (0,0)focus: $( 0 , - 3 )$
B)vertex: (0,0)focus: $( 0,3 )$
C)vertex: $$( 3,0 )$$ focus: (0,0)
D)vertex: $$( - 3,0 )$$ focus: (0,0)
E)vertex: (0,0)focus: $( 3,0 )$

Question

Quizplus · Accepted Answer

The given equation $x^2 + 12y = 0$ can be rewritten as $x^2 = -12y$, indicating a parabola that opens downwards. The vertex form of a parabola that opens vertically is $x^2 = 4py$, where $p$ is the distance from the vertex to the focus. Comparing the given equation to the vertex form, we find $4p = -12$, so $p = -3$. This means the focus is 3 units below the vertex, which is at the origin (0,0). Therefore, the focus is at $(0, -3)$.

Find the Vertex and Focus of the Parabola x2+12y=0x ^ { 2 } + 12 y = 0x2+12y=0

Find the Vertex and Focus of the Parabola $x ^ { 2 } + 12 y = 0$