In short

A parabola is the locus of all points in the plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). When the vertex sits at the origin and the axis lies along the x-axis, its equation is y^2 = 4ax. The number a controls how wide the curve opens and where the focus sits. The four standard orientations, the anatomy of the curve (vertex, axis, focus, directrix, latus rectum), and the ability to read all of them from the equation are the foundation of every parabola problem.

Stand at one end of a cricket ground and watch a fielder throw the ball back to the wicket-keeper. The ball traces a smooth arc through the air — rising, slowing at the top, then falling. That arc is a parabola. Every projectile in uniform gravity follows one. Every satellite dish is shaped like one. Every headlight reflector, every radar antenna, every solar concentrator — parabolas, all of them.

What makes this one curve so universally useful? It has a defining geometric property: every point on the curve is the same distance from a fixed point (called the focus) and a fixed line (called the directrix). This single property — one distance equals another — is responsible for everything. A satellite dish collects signals at its focus because parallel waves bouncing off a parabolic surface all converge to that one point. A headlight sends out a parallel beam because a bulb at the focus reflects light in parallel rays off the parabolic reflector. The geometry is doing the physics.

Before any of that engineering, though, you need the equation. And the equation comes from the definition.

Building the equation from the definition

Here is the setup. You have a fixed point S (the focus) and a fixed line d (the directrix). A point P moves in the plane such that its distance from S always equals its distance from d. The path traced by P is the parabola.

To turn this geometric condition into an equation, you need a coordinate system. The cleverest choice is to place the geometry so the algebra comes out clean. Put the vertex — the point on the parabola closest to the directrix — at the origin O(0, 0). Let the axis of the parabola (the line through the vertex and the focus) run along the positive x-axis. The focus then sits at S(a, 0) for some positive number a, and the directrix is the vertical line x = -a.

Why is the vertex halfway between the focus and the directrix? Because the vertex itself is on the parabola, so its distance to the focus must equal its distance to the directrix. The distance from O(0,0) to S(a, 0) is a, and the distance from O(0,0) to the line x = -a is also a. Equal — so the vertex lies on the curve. And being the closest point on the curve to the directrix, it sits exactly at the midpoint.

The parabola $y^2 = 4ax$ with $a = 2$. The vertex is at the origin, the focus is at $(2, 0)$, and the directrix is the vertical line $x = -2$. For any point $P$ on the curve, the distance $PS$ to the focus equals the perpendicular distance $PM$ to the directrix.

Now take a general point P(x, y) on the parabola. By the definition, the distance from P to the focus S(a, 0) equals the distance from P to the directrix x = -a.

The distance from P to the focus is:

PS = \sqrt{(x - a)^2 + y^2}

The distance from P to the directrix x = -a is the perpendicular (horizontal) distance from the point to the vertical line:

PM = |x - (-a)| = |x + a|

Set them equal:

\sqrt{(x - a)^2 + y^2} = |x + a|

Square both sides to remove the square root:

(x - a)^2 + y^2 = (x + a)^2

Why: squaring is safe here because both sides are non-negative (they are distances). No extraneous roots are introduced.

Expand both sides:

x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2

The x^2 and a^2 terms cancel from both sides:

y^2 - 2ax = 2ax
y^2 = 4ax

Why: the cross terms that survive after cancellation are -2ax on the left and +2ax on the right, giving y^2 = 4ax. All the squared terms disappeared — that is the beauty of placing the vertex at the origin.

Five lines of algebra, and the equation of the parabola drops out. The definition — "distance to focus equals distance to directrix" — compresses into three symbols: y^2 = 4ax.

Standard form of a right-opening parabola

The equation of a parabola with vertex at the origin, axis along the positive x-axis, focus at (a, 0), and directrix x = -a is

y^2 = 4ax \qquad (a > 0)

Notice that y is squared, not x. This is because the axis of the parabola lies along the x-axis, and the curve opens to the right — for each positive x, there are two values of y (one positive, one negative), and the curve spreads symmetrically above and below the axis. If x were squared instead, the axis would be vertical and the parabola would open upward or downward — which is the next story.

The left-opening parabola

What if you place the focus to the left of the vertex — at S(-a, 0) — with the directrix at x = a? Repeat the same derivation. For a point P(x, y):

PS = \sqrt{(x + a)^2 + y^2}, \qquad PM = |x - a|

Square both sides of PS = PM:

(x + a)^2 + y^2 = (x - a)^2
x^2 + 2ax + a^2 + y^2 = x^2 - 2ax + a^2
y^2 = -4ax

The only difference from the right-opening case is the sign: y^2 = -4ax instead of y^2 = 4ax. The negative sign tells you the curve opens in the negative x-direction — to the left.

The four orientations

By placing the focus on each of the four semi-axes (positive x, negative x, positive y, negative y), you get four standard parabolas. They are all the same shape; only the direction differs.

Right-opening: y^2 = 4ax — focus at (a, 0), directrix x = -a.

Left-opening: y^2 = -4ax — focus at (-a, 0), directrix x = a.

Upward-opening: x^2 = 4ay — focus at (0, a), directrix y = -a.

Downward-opening: x^2 = -4ay — focus at (0, -a), directrix y = a.

The upward and downward forms come from the same derivation but with the roles of x and y swapped. For instance, when the focus is at (0, a) and the directrix is y = -a, the locus condition gives:

\sqrt{x^2 + (y - a)^2} = |y + a|

Squaring: x^2 + y^2 - 2ay + a^2 = y^2 + 2ay + a^2, so x^2 = 4ay.

All four standard parabolas with $a = 1$, plotted on the same axes. The solid black curve opens right ($y^2 = 4x$, focus $S_1$), the solid red opens left ($y^2 = -4x$, focus $S_2$), the dashed dark curve opens upward ($x^2 = 4y$, focus $S_3$), and the dashed light curve opens downward ($x^2 = -4y$, focus $S_4$).

The pattern is simple enough to remember without memorising. When y is squared, the axis is horizontal — the parabola opens left or right. When x is squared, the axis is vertical — the parabola opens up or down. The sign on the right side tells you the direction: positive means the curve opens toward positive values of the non-squared variable, negative means it opens toward negative values.

A useful mnemonic: in y^2 = 4ax, the variable that is not squared (x) tells you which axis the parabola opens along. The sign of the coefficient of that variable tells you which way.

The anatomy of a parabola

Every parabola problem uses the same vocabulary. Here are the five terms you need.

Vertex. The point where the parabola turns. In standard form, the vertex is at the origin (0, 0). It is the point on the parabola closest to the directrix. Geometrically, it is the "tip" of the curve.

Focus. The fixed point S in the definition. For y^2 = 4ax, the focus is at (a, 0). The focus lies inside the curve, on the axis of symmetry, at a distance a from the vertex. The parabola "cups around" the focus.

Directrix. The fixed line in the definition. For y^2 = 4ax, the directrix is the vertical line x = -a. It is on the opposite side of the vertex from the focus, at the same distance a. The directrix lies entirely outside the curve.

Axis. The line of symmetry of the parabola, passing through the vertex and the focus. For y^2 = 4ax, the axis is the x-axis (i.e., y = 0). The parabola is symmetric about this line: if (x, y) is on the parabola, so is (x, -y).

Latus rectum. The chord of the parabola that passes through the focus and is perpendicular to the axis. It is a direct measure of how "wide" the parabola is at the level of the focus.

Deriving the latus rectum length

To find the endpoints of the latus rectum for y^2 = 4ax, substitute x = a (the x-coordinate of the focus) into the equation:

y^2 = 4a \cdot a = 4a^2 \implies y = \pm 2a

So the endpoints are (a, 2a) and (a, -2a). These are two points on the parabola that lie directly above and below the focus.

The length of the latus rectum is the distance between these endpoints:

\text{Length} = 2a - (-2a) = 4a

Each half of the latus rectum — from the focus to one endpoint — has length 2a. This half-length 2a is called the semi-latus rectum.

The parabola $y^2 = 8x$ (so $a = 2$) with all its parts labelled. The vertex is at the origin, the focus at $(2, 0)$, the directrix at $x = -2$, and the latus rectum is the red vertical chord through the focus from $(2, 4)$ to $(2, -4)$, with length $4a = 8$.

The latus rectum deserves a moment of attention. Its length 4a is four times the distance from the vertex to the focus. A large a means the focus is far from the vertex and the parabola opens wide (more "spread out"); a small a means the focus is close and the parabola is narrow (more "pinched"). The single parameter a controls the entire shape of the parabola — it fixes the focus, the directrix, and how quickly the curve spreads. Two parabolas with the same value of a are identical curves (just possibly in different orientations or positions).

The complete reference table

Here is the reference for all four orientations. Every entry follows from the focus-directrix derivation, with the focus placed in the appropriate direction.

Equation Focus Directrix Axis Opens toward Latus rectum
y^2 = 4ax (a, 0) x = -a y = 0 Right 4a
y^2 = -4ax (-a, 0) x = a y = 0 Left 4a
x^2 = 4ay (0, a) y = -a x = 0 Upward 4a
x^2 = -4ay (0, -a) y = a x = 0 Downward 4a

In all four cases, a > 0. The sign in the equation encodes the direction. The latus rectum length is always 4a.

How to read a from an equation

This is a mechanical step, but getting it right matters — most errors in parabola problems come from misreading a. The method is always the same: match the given equation to the appropriate standard form.

Example. Given y^2 = 12x, compare with y^2 = 4ax:

4a = 12 \implies a = 3

Since the coefficient of x is positive and y is squared, the parabola opens to the right. The focus is at (3, 0), the directrix is x = -3, and the latus rectum has length 12.

Example. Given x^2 = -20y, compare with x^2 = -4ay:

4a = 20 \implies a = 5

The negative sign and the fact that x is squared tell you the parabola opens downward. The focus is at (0, -5), the directrix is y = 5, and the latus rectum has length 20.

Example. Given y^2 + 24x = 0, rearrange first: y^2 = -24x. Compare with y^2 = -4ax:

4a = 24 \implies a = 6

This opens to the left. Focus at (-6, 0), directrix x = 6.

The first step is always: isolate the squared term on one side, put everything else on the other, and match.

How the value of a affects the shape

To build intuition for what different values of a look like, consider three right-opening parabolas: y^2 = 4x (so a = 1), y^2 = 8x (so a = 2), and y^2 = 16x (so a = 4).

At x = 4: for a = 1, y = \pm 4; for a = 2, y = \pm 4\sqrt{2} \approx \pm 5.66; for a = 4, y = \pm 8. The larger a is, the wider the parabola spreads at the same distance from the vertex.

Three right-opening parabolas with different values of $a$. The narrowest curve (grey) has $a = 1$, the middle one (black) has $a = 2$, and the widest (red) has $a = 4$. A larger $a$ pushes the focus farther from the vertex and makes the curve open more gently.

The analogy is a bowl: a large a is a wide, shallow bowl, and a small a is a narrow, deep bowl. In satellite dish terms, a larger a means a shallower dish with the receiver farther from the surface.

Worked examples

Example 1: Finding the equation from the focus

A parabola has its vertex at the origin and its focus at (0, 3). Find its equation, directrix, the endpoints and length of the latus rectum, and verify the focus-directrix property at one endpoint.

Step 1. Identify the orientation. The focus is at (0, 3) — on the positive y-axis. So the parabola opens upward and the standard form is x^2 = 4ay.

Why: the focus lies on the axis of symmetry. Since it is above the origin on the y-axis, the axis is vertical and the parabola opens upward.

Step 2. Read off a. The focus is at (0, a), and here that is (0, 3), so a = 3.

Why: the distance from the vertex to the focus is a, by definition of the standard form.

Step 3. Write the equation.

x^2 = 4(3)y = 12y

Why: substitute a = 3 into x^2 = 4ay.

Step 4. Find the directrix. For an upward-opening parabola, the directrix is y = -a:

\text{directrix: } y = -3

Why: the directrix is always on the opposite side of the vertex from the focus, at the same distance a.

Step 5. Find the latus rectum. The latus rectum passes through the focus perpendicular to the axis. The focus is at (0, 3) and the axis is vertical, so the latus rectum is horizontal at y = 3. Substitute y = 3 into x^2 = 12y: x^2 = 36, so x = \pm 6. The endpoints are (-6, 3) and (6, 3), and the length is 12 = 4a.

Why: substituting the y-coordinate of the focus into the equation gives the x-coordinates of the latus rectum endpoints.

Now verify: the endpoint (6, 3) should be equidistant from the focus (0, 3) and the directrix y = -3. Distance to focus: \sqrt{36 + 0} = 6. Distance to directrix: 3 - (-3) = 6. Equal.

Result: The equation is x^2 = 12y, the directrix is y = -3, the latus rectum endpoints are (-6, 3) and (6, 3) with length 12.

The parabola $x^2 = 12y$ opens upward with vertex at the origin and focus at $(0, 3)$. The directrix is the dashed line $y = -3$. The latus rectum (red segment) passes through the focus from $(-6, 3)$ to $(6, 3)$, confirming its length is $12$.

The graph confirms the algebra: the curve cups upward around the focus, the latus rectum spans the width of the parabola at the focus, and both endpoints sit on the curve. The parabola is symmetric about the y-axis — for every point (x, y) on the curve, (-x, y) is also on the curve.

Example 2: Equation of a parabola with given latus rectum

The latus rectum of a parabola is 16 units long, the vertex is at the origin, and the parabola opens to the left. Find the equation, the focus, the directrix, and verify by checking two points.

Step 1. Find a from the latus rectum. The latus rectum has length 4a:

4a = 16 \implies a = 4

Why: the length of the latus rectum is always 4a, regardless of orientation. This is the most direct way to recover a when you know the latus rectum length.

Step 2. Identify the standard form. The parabola opens to the left, so the form is y^2 = -4ax.

Why: opening left means the axis is horizontal (so y is squared) and the curve extends in the negative x direction (so the sign is negative).

Step 3. Write the equation.

y^2 = -4(4)x = -16x

Why: substitute a = 4 into y^2 = -4ax.

Step 4. Find the focus and directrix. For y^2 = -4ax, the focus is at (-a, 0) = (-4, 0) and the directrix is x = a = 4.

Why: in a left-opening parabola, the focus is to the left of the vertex and the directrix is to the right, each at distance a from the vertex.

Step 5. Verify the endpoints. Substitute x = -4 into y^2 = -16x: y^2 = -16(-4) = 64, so y = \pm 8. The latus rectum endpoints are (-4, 8) and (-4, -8), and the distance between them is 16.

Check (−4, 8): distance to focus (-4, 0) is \sqrt{0 + 64} = 8. Distance to directrix x = 4 is |-4 - 4| = 8. Equal.

Result: The equation is y^2 = -16x, the focus is at (-4, 0), and the directrix is x = 4.

The parabola $y^2 = -16x$ opens to the left. The focus at $(-4, 0)$ and the directrix $x = 4$ are on opposite sides of the vertex. The latus rectum stretches from $(-4, -8)$ to $(-4, 8)$, confirming its length is $16$.

The picture shows the curve opening to the left, with the focus sitting inside the "cup" of the parabola. The directrix is a vertical line on the right, equidistant from the vertex as the focus. Every point on the left-opening curve has negative x-coordinates (except the vertex at the origin) — this is visible in the graph and follows from y^2 = -16x requiring x \leq 0.

Common confusions

A few things students reliably get wrong about parabolas the first time they meet them.

Going deeper

If you came here to learn the definition, standard forms, and terminology of a parabola, you have it — you can stop here. The rest of this section is for readers who want to understand the reflective property, shifted parabolas, and the place of the parabola among the conic sections.

The reflective property

The parabola's most celebrated physical property is this: a ray traveling parallel to the axis and hitting the parabola will reflect through the focus. Conversely, a ray emanating from the focus will reflect off the parabola and travel parallel to the axis.

This is why every satellite dish, car headlight, and solar concentrator is shaped as a parabola. A satellite dish collects incoming parallel signals and redirects them all to the focus, where the receiver sits. A headlight bulb placed at the focus sends its light outward as a parallel beam, rather than scattering in all directions.

The proof rests on the tangent line at each point. At any point P on the parabola, the tangent bisects the angle between the line from P to the focus (the focal radius) and the line from P perpendicular to the directrix. Since the perpendicular to the directrix is parallel to the axis, the angle of incidence (measured from the tangent) equals the angle of reflection — exactly the condition for a mirror reflection. The full proof using coordinates is given in Parabola — Tangent and Normal.

Bhaskara II, the 12th-century Indian mathematician, studied the properties of conic curves in his work Siddhanta Shiromani, and the parabolic shape was used in the design of fire mirrors (burning mirrors) in several ancient civilisations — a concave parabolic mirror can concentrate sunlight at the focus with enough intensity to start a fire.

Shifted parabolas

The standard form places the vertex at the origin. If the vertex is shifted to a point (h, k), the equation becomes:

(y - k)^2 = 4a(x - h) \qquad \text{(right-opening)}
(y - k)^2 = -4a(x - h) \qquad \text{(left-opening)}
(x - h)^2 = 4a(y - k) \qquad \text{(upward-opening)}
(x - h)^2 = -4a(y - k) \qquad \text{(downward-opening)}

The shape of the curve does not change — only its position does. The focus, directrix, and latus rectum all shift by the same amount. For instance, in the right-opening shifted form, the focus moves to (h + a, k) and the directrix becomes x = h - a.

When you encounter a parabola equation that does not match a standard form directly — say y^2 - 6y - 8x + 1 = 0 — the first step is to complete the square on the variable that is squared, group terms, and identify (h, k) and a. For this example: y^2 - 6y = 8x - 1, then (y - 3)^2 - 9 = 8x - 1, so (y - 3)^2 = 8(x + 1). This gives k = 3, h = -1, 4a = 8 so a = 2. Vertex at (-1, 3), focus at (1, 3), directrix x = -3.

The eccentricity viewpoint

Every conic section — circle, ellipse, parabola, hyperbola — can be defined as the locus of points where the ratio of distance-to-focus to distance-to-directrix is a constant e, called the eccentricity.

\frac{\text{distance to focus}}{\text{distance to directrix}} = e

For a parabola, e = 1 — the two distances are always equal, which is exactly the definition you started with. For an ellipse, e < 1 (the focus is closer than the directrix). For a hyperbola, e > 1 (the focus is farther than the directrix). The parabola sits at the exact boundary between the ellipse and the hyperbola — the critical case where the ratio is exactly one. In the hierarchy of conics, the parabola is the "just right" curve.

This is also visible in the cone-slicing picture: cut a cone with a plane parallel to its slant edge, and the cross-section is a parabola. Tilt the plane one way for an ellipse, the other way for a hyperbola. The parabola is the boundary angle.

Where this leads next

You now have the definition, standard forms, and vocabulary of the parabola. The next articles build on this foundation: