Q.Find the equation of the parabola that satisfies the given conditions: Focus $(0, -3)$; directrix $y = 3$.

Concept: Standard form of a parabola with vertical axis.

A parabola is the locus of points equidistant from the focus and directrix. Here the focus is at $(0, -3)$ and the directrix is the horizontal line $y = 3$.

The vertex lies midway between focus and directrix:

$\text{Vertex} = \left(0, \frac{-3 + 3}{2}\right) = (0, 0)$ …

The defining property of a parabola is that every point on it sits at equal distances from a fixed point (the focus) and a fixed line (the directrix). This geometric definition gives us a direct path to the equation.

Here the focus is at $(0, -3)$, below the $x$-axis, and the directrix is the horizontal line $y = 3$, above it. The parabola must open downward, wrapping around the focus and away from the directrix.

The vertex of a parabola lies exactly halfway between the focus and directrix. The focus is at $y = -3$ and the directrix at $y = 3$, so the vertex sits at:

Since the focus has $x$-coordinate $0$, the axis of symmetry is the $y$-axis, placing the vertex at $(0, 0)$.

Now we build the equation from the definition. Let $(x, y)$ be any point on the parabola.

1. Distance from $(x, y)$ to the focus $(0, -3)$:

2. Distance from $(x, y)$ to the directrix $y = 3$:

   The perpendicular distance from a point to a horizontal line $y = k$ is simply $|y - k|$. Here:

3. Set the distances equal:

   $\sqrt{x^2 + (y + 3)^2} = |y - 3|$ …