Parabola — Equation, Focus, Directrix & Graph

When a problem hands you an equation like $(x-2)^2 = 12(y+1)$ and asks for the vertex, focus, directrix, and opening direction, you need a reliable procedure rather than a guess. This is that procedure, built on one defining fact: a parabola is the set of all points equally far from a fixed focus and a fixed line, the directrix.

When to use this method

Reach for this matching procedure whenever the equation is already in (or can be put into) one of the standard forms below. The U-shaped picture is only a hint; the procedure works from the algebra, so it handles parabolas that open up, down, left, or right, and parabolas whose vertex is not at the origin.

The standard forms, with vertex at $(h,k)$:

For the vertical form the focus is $(h, k+p)$ and the directrix is $y = k-p$. For the horizontal form the focus is $(h+p, k)$ and the directrix is $x = h-p$. With vertex at the origin these reduce to $x^2=4py$ (focus $(0,p)$, directrix $y=-p$) and $y^2=4px$ (focus $(p,0)$, directrix $x=-p$).

The steps

1. Find the form. Match the equation to $(x-h)^2 = 4p(y-k)$ or $(y-k)^2 = 4p(x-h)$ so you know whether the parabola is vertical or horizontal.
2. Read the vertex. The vertex is $(h,k)$; if the equation is unshifted, it is the origin.
3. Compute $p$. Compare against $4p$, not $p$ — the coefficient next to the variable equals $4p$.
4. Locate focus and directrix. Use the sign of $p$ to place them on the axis of symmetry.
5. Sketch the opening. Plot the vertex and focus, draw the directrix, and sketch the curve opening toward the focus.

The vertex always lies halfway between focus and directrix along the axis of symmetry, and the curve always opens toward the focus and away from the directrix.

The whole procedure on one example

Consider

Form: matches $(x-h)^2 = 4p(y-k)$, so it is vertical. Vertex: $h=2,\ k=-1$, giving $(2,-1)$. Compute $p$: $4p = 12$, so $p = 3$. Focus and directrix: since $p>0$ it opens upward; focus $(2,\,-1+3) = (2,2)$ and directrix $y = -1-3 = -4$. Sketch: axis of symmetry $x=2$, vertex $(2,-1)$, opening upward toward $(2,2)$. Reading it back as a list:

• Vertex: $(2, -1)$
• Axis of symmetry: $x = 2$
• Opening: upward, since $p > 0$
• Focus: $(2, 2)$
• Directrix: $y = -4$

Where each step tends to stall, and how to check

Step 1 (form): if you cannot tell vertical from horizontal, look at which variable is squared. Squared $x$ means vertical; squared $y$ means horizontal. Swapping these gives the wrong focus and directrix.

Step 3 (compute $p$): the most frequent slip is reading the coefficient as $p$ instead of $4p$. In $(x-h)^2 = 12(y-k)$, $4p=12$ gives $p=3$, not $p=12$.

Step 4 (sign): a negative $p$ opens the parabola down or left, not up or right. The sign controls direction.

Self-check: after sketching, confirm the focus lies on the side where the curve opens, and that the vertex is at $(0,0)$ only when the form is genuinely unshifted. If you forget the formulas, rebuild them from the geometry — equal distance to focus and directrix, vertex in the middle, opening toward the focus.

Try the full procedure on $(y + 3)^2 = -8(x - 1)$: identify the form, read the vertex, find $p$ from $4p=-8$, place the focus and directrix, and confirm the opening direction before sketching. Parabolas appear in coordinate geometry, quadratic graphs, and conic sections, and in idealized projectile motion; their reflection property (rays parallel to the axis pass through the focus) is why parabolic dishes, reflectors, and mirrors exist.