The given equation is:

$x^2 + 12y = 0$

First, we rearrange the equation into standard form by isolating $y$:

$x^2 = -12y$

This is a parabola that opens downward since the coefficient of $y$ is negative. The standard form of a parabola that opens vertically is:

$x^2 = 4p y$

Here, comparing $x^2 = -12y$ with $x^2 = 4p y$, we can see that $4p = -12$. Solving for $p$:

$p = \frac{-12}{4} = -3$

Vertex:

The vertex of the parabola is at the origin, $(0, 0)$, since the equation is in the form $x^2 = 4p(y - k)$ and there is no shift in either the $x$- or $y$-direction.

Focus:

The focus of a parabola is located at a distance $p$ from the vertex along the axis of symmetry. Since $p = -3$, the focus is 3 units below the vertex (because $p$ is negative and the parabola opens downward).

Thus, the focus is at:

$(0, -3)$

Summary:

• Vertex: $(0, 0)$
• Focus: $(0, -3)$

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the directrix of the given parabola?
2. How do you find the axis of symmetry of a parabola?
3. What would be the equation if the parabola opened to the right?
4. How do you determine the direction in which a parabola opens?
5. How does the value of $p$ affect the shape of the parabola?

Tip: The focus and directrix of a parabola always lie on opposite sides of the vertex.