Find the vertex and focus of the parabola. $x ^ { 2 } + 28 y = 0$
A)vertex: (-7 , 0) focus: (0, 0)
B)vertex: (7, 0) focus: (0, 0)
C)vertex: (0, 0) focus: (-7, 0)
D)vertex: (0, 0) focus: (0, -7)
E)vertex: (0, 0) focus: (0, 7)

Question

Quizplus · Accepted Answer

The given equation is $x^2 + 28y = 0$, which can be rewritten as $x^2 = -28y$. This is 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 that $4p = -28$, so $p = -7$. Since $p$ is negative, the parabola opens downwards, and the focus is below the vertex. The vertex is at the origin (0, 0), and the focus is $p = -7$ units below the vertex, so the focus is at (0, -7).

Find the Vertex and Focus of the Parabola x2+28y=0x ^ { 2 } + 28 y = 0x2+28y=0

Find the Vertex and Focus of the Parabola $x ^ { 2 } + 28 y = 0$