Parabola — standard forms (4 orientations), focus, directrix, latus rectum, axis: Worked examples

The anatomy picture (read this first)

Every example below moves the same four objects around: the focus (a point), the directrix (a line), the vertex (the midpoint between them), and the latus rectum — the short chord through the focus. Before any algebra, look at how they sit together on one right-opening parabola:

Figure — Parabola — standard forms (4 orientations), focus, directrix, latus rectum, axis

In the figure, notice these things that make the definitions visual:

the focus sits inside the bowl of the curve, the directrix sits outside it;
the vertex is exactly halfway between them (the two little equal ticks, each of length $a$ );
the axis runs horizontally through focus and vertex — fold along it and the top and bottom match;
for the marked point $P$ , the two red dashed segments — to the focus and straight across to the directrix — are equal in length. That equality is the parabola.

The scenario matrix

Now let's list every kind of situation a parabola question can be. Each row is a "cell" we must hit at least once.

Cell	What makes it different	Covered by
A. Right-opening ( $y^{2} = + 4 a x$ )	positive coeff, horizontal axis	Ex 1
B. Left-opening ( $y^{2} = - 4 a x$ )	negative coeff, focus on $- x$ side	Ex 2
C. Up-opening ( $x^{2} = + 4 a y$ )	squared $x$ , vertical axis	Ex 3
D. Down-opening ( $x^{2} = - 4 a y$ )	negative coeff, focus below	Ex 3
E. Build from focus + directrix	reverse direction, pick the form	Ex 4
F. Build from definition (raw)	no standard form assumed	Ex 5
G. Point-on-curve check	is a given point on the parabola?	Ex 6
H. Degenerate / limiting ( $a \to 0$ )	focus meets directrix, curve → line	Ex 7
I. Real-world word problem	dish / thrown ball, units matter	Ex 8
J. Exam twist — horizontal shift	vertex not at origin, opens sideways	Ex 9
K. Exam twist — vertical shift, focus in a quadrant	shifted vertex, opens down, focus with $x < 0, y < 0$	Ex 10

We also make sure all four quadrants of the focus appear, not just the axes: Ex 8 puts the focus in Quadrant I ( $x > 0, y > 0$ ), Ex 9's focus lands in Quadrant I too, while Ex 10 places the focus in Quadrant III ( $x < 0, y < 0$ ) and, along the way, shows a companion point in Quadrants II and IV.

Ex 1 — Cell A: right-opening

Worked example Find the focus, directrix, axis and latus rectum of

$y^{2} = 20 x$ . Forecast: guess first — is $a$ equal to $20$ ? Which way does it open? Jot your guess.

Step 1 — Match the template $y^{2} = 4 a x$ . The equation has $y^{2}$ on the left and a positive multiple of $x$ on the right, so it fits $y^{2} = 4 a x$ exactly. Why this step? Recognising the template tells us the orientation before any arithmetic: $y^{2} \Rightarrow$ horizontal axis, positive coefficient $\Rightarrow$ opens right.

Step 2 — Read off $a$ . $4 a = 20 \Rightarrow a = 5$ . Why this step? The coefficient is written as $4 a$ , so $a$ (the focal distance we defined above) is one-quarter of it — never the whole number. (This " $a$ vs $4 a$ " trap is spelled out in the parent's mistakes list: see §6 of the parent note.)

Step 3 — Place the parts. Focus $= (a, 0) = (5, 0)$ . Directrix $= x = - a = - 5$ . Axis $= y = 0$ . Latus rectum (LR) $= 4 a = 20$ . Why this step? Each part comes straight from the standard-form table; the focus is on the same side as the opening (right).

Verify: Take the vertex $(0, 0)$ . Distance to focus $= 5$ . Distance to directrix $x = - 5$ is $∣0 - (- 5) ∣ = 5$ . Equal ✓ — the vertex obeys the defining law.

Ex 2 — Cell B: left-opening (negative coefficient)

Worked example Analyse

$y^{2} = - 16 x$ . Forecast: which side is the focus on now — left or right?

Step 1 — Match $y^{2} = - 4 a x$ . Still $y^{2}$ , but the coefficient is negative, so we use the left-opening template. Why this step? The minus sign is the only thing that changed from Ex 1, and its whole job is to flip the direction.

Step 2 — Extract $a$ from the magnitude. $4 a = 16 \Rightarrow a = 4$ . We keep $a > 0$ (convention) and let the sign carry the direction. Why this step? If we let $a$ be negative, " $a$ " would no longer mean the focal distance. We separate size ( $a$ ) from direction (the sign).

Step 3 — Place the parts, flipping signs. Focus $= (- a, 0) = (- 4, 0)$ . Directrix $= x = + a = 4$ . Axis $= y = 0$ . LR $= 4 a = 16$ . Why this step? Opening left means the focus moves to $- x$ , and the directrix to the opposite side $+ x$ .

Verify: Vertex $(0, 0)$ : distance to focus $∣ - 4 - 0∣ = 4$ ; distance to directrix $x = 4$ is $∣0 - 4∣ = 4$ . Equal ✓.

Ex 3 — Cells C & D: up vs down (squared $x$ )

Worked example Contrast

$x^{2} = 8 y$ (call it P) with $x^{2} = - 8 y$ (call it Q). Forecast: both have $x^{2}$ . Which opens up and which opens down? Where do their foci sit?

Step 1 — Match templates. P fits $x^{2} = 4 a y$ (positive) → opens up. Q fits $x^{2} = - 4 a y$ (negative) → opens down. Why this step? When $x$ is squared, the non-squared variable $y$ names the axis, so both are vertical. The sign then picks up vs down. (This kills the "there's an $x$ so it opens sideways" mistake.)

Step 2 — Same $a$ for both. $4 a = 8 \Rightarrow a = 2$ in each. Why this step? The magnitude of the coefficient is identical, so both have the same "width"; only direction differs.

Step 3 — Foci and directrices. P: Focus $(0, a) = (0, 2)$ , directrix $y = - 2$ . Axis $x = 0$ . LR $= 8$ . Q: Focus $(0, - a) = (0, - 2)$ , directrix $y = 2$ . Axis $x = 0$ . LR $= 8$ . Why this step? Focus always sits on the opening side; directrix opposite.

Verify (using the defining property, like Ex 1 & 2): Test P's vertex $(0, 0)$ : distance to focus $(0, 2)$ is $2$ ; distance to directrix $y = - 2$ is $∣0 - (- 2) ∣ = 2$ — equal ✓. Test Q's vertex $(0, 0)$ : distance to focus $(0, - 2)$ is $2$ ; distance to directrix $y = 2$ is $∣0 - 2∣ = 2$ — equal ✓. Both obey the parabola law.

The two curves and their mirror-image foci/directrices sit back-to-back here; look at how the red up-parabola cups its focus above the axis while the black down-parabola cups its focus below:

Ex 4 — Cell E: build from focus + directrix

Worked example Find the equation with focus

$(3, 0)$ and directrix $x = - 3$ . Forecast: which of the four forms will it be?

Step 1 — Locate the vertex. Vertex is the midpoint of the focus and its foot on the directrix: midpoint of $(3, 0)$ and $(- 3, 0)$ is $(0, 0)$ . Why this step? The vertex must sit on the parabola and on the axis, exactly halfway — that's the definition applied at the closest point.

Step 2 — Choose orientation. Focus is on the right of the vertex, directrix on the left → opens right → form $y^{2} = 4 a x$ . Why this step? The parabola always wraps around its focus, so it opens toward the focus side.

Step 3 — Find $a$ and write it. Focal distance $= ∣3 - 0∣ = 3$ , so $a = 3$ . Equation: $y^{2} = 4 (3) x = 12 x$ . Why this step? $a$ is just the vertex-to-focus distance.

Verify: Directrix should be $x = - a = - 3$ ✓ (matches the given). LR $= 4 a = 12$ .

Ex 5 — Cell F: straight from the definition (no template)

Worked example Using only "distance to focus = distance to directrix", find the parabola with focus

$(0, - 5)$ and directrix $y = 5$ . Forecast: we won't peek at any table — predict the final equation.

Step 1 — Write both distances for a general point $P = (x, y)$ . Distance to focus (by the distance formula): $(x - 0)^{2} + (y + 5)^{2}$ . Distance to directrix $y = 5$ (perpendicular gap): $∣ y - 5∣$ . Why this step? This is the raw definition — every standard form is this sentence in coordinates.

Step 2 — Set them equal and square. $x^{2} + (y + 5)^{2} = (y - 5)^{2}$ Why square? To remove the square root and the absolute value in one clean move — both sides are non-negative, so squaring is safe and reversible here.

Step 3 — Expand and simplify. $x^{2} + y^{2} + 10 y + 25 = y^{2} - 10 y + 25 ⟹ x^{2} = - 20 y .$ Why this step? The $y^{2}$ and $25$ cancel, leaving a pure standard form we derived, not assumed.

Verify: $x^{2} = - 20 y = - 4 a y \Rightarrow 4 a = 20 \Rightarrow a = 5$ ; focus $(0, - 5)$ ✓, directrix $y = 5$ ✓, opens down as expected.

Ex 6 — Cell G: is a point on the parabola?

Worked example Does the point

$(3, - 6)$ lie on $y^{2} = 12 x$ ? If yes, find its distance to the focus. Forecast: guess yes or no before computing.

Step 1 — Test membership. LHS $= y^{2} = (- 6)^{2} = 36$ . RHS $= 12 x = 12 (3) = 36$ . Equal → yes, it lies on the curve. Why this step? A point is on the curve exactly when its coordinates satisfy the equation.

Step 2 — Find the focus. $4 a = 12 \Rightarrow a = 3$ , focus $(3, 0)$ , directrix $x = - 3$ . Why this step? We need the focus to use the defining distance property as a shortcut.

Step 3 — Focal distance the smart way. By the definition, distance to focus = distance to directrix. For a point $(x, y)$ the directrix is the vertical line $x = - a = - 3$ , so the perpendicular gap to it is $∣ x - (- 3) ∣ = ∣ x + 3∣$ . Since our point has $x = 3 > 0$ , this is $x + 3 = 3 + 3 = 6$ . Why this step? Instead of computing the square-root distance to the focus, we swap it (legally, by the parabola's defining equality) for the much simpler horizontal gap to the directrix. The expression $x + a$ comes directly from the distance to the line $x = - a$ : subtract the line's $x$ -value $(- a)$ from the point's $x$ -value, giving $x + a$ (no absolute value needed once $x \geq 0$ , since the curve lives on the right of the vertex).

Verify: Direct distance $(3 - 3)^{2} + (- 6 - 0)^{2} = 36 = 6$ ✓ — matches the shortcut.

Ex 7 — Cell H: the degenerate / limiting case $a \to 0$

Worked example What happens to

$y^{2} = 4 a x$ as $a \to 0^{+}$ ? Is it still a parabola? Forecast: picture the focus at $(a, 0)$ and directrix at $x = - a$ . What do they do as $a$ shrinks?

Step 1 — Watch the pieces move. Focus $(a, 0) \to (0, 0)$ and directrix $x = - a \to x = 0$ . The focus lands on the directrix. Why this step? The definition requires the directrix not to pass through the focus. As $a \to 0$ that condition breaks — so we're leaving parabola-land.

Step 2 — Look at the equation. $y^{2} = 4 a x$ with $a = 0$ gives $y^{2} = 0$ , i.e. $y = 0$ . Why this step? Substituting the limiting value shows the curve algebraically.

Step 3 — Interpret. The parabola collapses onto its own axis — a single line (a degenerate conic), not a parabola. Why this step? Every point satisfying $y = 0$ is equidistant (both distances become $∣ x ∣$ ) — the "curve" flattens completely.

Verify: For tiny $a = 0.001$ , the latus rectum $= 4 a = 0.004$ — a nearly invisible sliver; the curve hugs the $x$ -axis. As $a \to 0$ this width $\to 0$ ✓.

The figure shows the curve tightening toward the axis as $a$ shrinks; the red limiting line is what remains when $a = 0$ :

Ex 8 — Cell I: real-world word problem (units!)

Worked example A satellite dish is a parabola. It is

$2 m$ wide across the rim and $0.5 m$ deep at the centre. Where should the receiver (at the focus) be placed? Forecast: estimate — closer to the vertex than $0.5 m$ , or further?

Step 1 — Set convenient coordinates. Vertex at origin, dish opening up: $x^{2} = 4 a y$ . The rim point is at half-width $x = 1 m$ , depth $y = 0.5 m$ . Why this step? Placing the vertex at the origin makes the standard form apply directly; half the width because $x$ is measured from the axis.

Step 2 — Solve for $a$ . $(1)^{2} = 4 a (0.5) \Rightarrow 1 = 2 a \Rightarrow a = 0.5 m$ . Why this step? The rim point is on the curve, so its coordinates satisfy the equation — one equation, one unknown $a$ .

Step 3 — Locate the focus. Focus $= (0, a) = (0, 0.5 m)$ — exactly $0.5 m$ above the vertex, sitting in Quadrant I ( $x = 0$ on the axis, $y > 0$ ). Why this step? For $x^{2} = 4 a y$ , the focus is a distance $a$ up the axis. Units: metres throughout ✓.

Verify: Check rim on curve: $x^{2} = 1$ , $4 a y = 4 (0.5) (0.5) = 1$ ✓. The receiver sits right at rim height — a physically sensible dish. (This is the same parabola that governs thrown-ball paths.)

Ex 9 — Cell J: exam twist (horizontal shift, vertex off origin)

Worked example Find the vertex, focus and directrix of

$(y - 2)^{2} = 8 (x + 1)$ . Forecast: the vertex is not at the origin. Where is it?

Step 1 — Read the shift. Compare with $(y - k)^{2} = 4 a (x - h)$ . Here $h = - 1$ , $k = 2$ , so the vertex is $(- 1, 2)$ . Why this step? Replacing $x \to x - h$ , $y \to y - k$ just slides the whole standard parabola so its vertex sits at $(h, k)$ — the shape is unchanged.

Step 2 — Extract $a$ . $4 a = 8 \Rightarrow a = 2$ . Positive coefficient, $y$ squared → opens right. Why this step? Same reading rule as always; the shift doesn't affect orientation or $a$ .

Step 3 — Shift the focus and directrix. In local (unshifted) coordinates: focus $(a, 0) = (2, 0)$ , directrix $x = - a = - 2$ . Shift back by adding $(h, k) = (- 1, 2)$ : Focus $= (2 - 1, 0 + 2) = (1, 2)$ (Quadrant I). Directrix $x = - 2 - 1 = - 3$ . Why this step? Everything moves rigidly with the vertex, so add the same offset to each feature.

Verify: Vertex $(- 1, 2)$ : distance to focus $(1, 2)$ is $(1 - (- 1))^{2} + (2 - 2)^{2} = 4 = 2$ ; distance to directrix $x = - 3$ is $∣ - 1 - (- 3) ∣ = 2$ . Equal ✓, and the vertex is midway between them ✓.

Ex 10 — Cell K: exam twist (vertical shift, focus in Quadrant III)

Worked example A parabola opens

downward with vertex $(- 2, - 1)$ and $a = 2$ . Find its equation, focus and directrix, and locate the two latus-rectum endpoints. Forecast: guess the quadrant of the focus before computing — up or down from the vertex, left or right?

Step 1 — Pick the shifted template. Opens down + $x$ squared → $(x - h)^{2} = - 4 a (y - k)$ with vertex $(h, k) = (- 2, - 1)$ . Why this step? Down-opening means $x$ is squared and the coefficient is negative; the shift puts the vertex at $(h, k)$ .

Step 2 — Insert $a = 2$ and write the equation. $4 a = 8$ , so $(x + 2)^{2} = - 8 (y + 1)$ . Why this step? Substituting $h = - 2, k = - 1$ turns $x - h$ into $x + 2$ and $y - k$ into $y + 1$ ; the minus sign fixes the downward opening.

Step 3 — Focus and directrix by shifting. Local focus for a down-parabola is $(0, - a) = (0, - 2)$ ; add $(h, k) = (- 2, - 1)$ : Focus $= (- 2, - 3)$ — that is $x < 0, y < 0$ , firmly in Quadrant III. Local directrix $y = a = 2$ ; shift: $y = 2 + (- 1) = 1$ . Why this step? Down-opening pushes the focus below the vertex; the directrix sits above.

Step 4 — Latus-rectum endpoints (both horizontal signs). LR $= 4 a = 8$ , so the endpoints lie $\pm \frac{4 a}{2} = \pm 4$ sideways from the focus: $(- 2 - 4, - 3) = (- 6, - 3)$ and $(- 2 + 4, - 3) = (2, - 3)$ . Why this step? The LR is horizontal (parallel to the directrix) and centred on the focus, so its ends are half its length either side.

Verify: Vertex $(- 2, - 1)$ : distance to focus $(- 2, - 3)$ is $0 + (- 1 - (- 3))^{2} = 4 = 2 = a$ ✓; distance to directrix $y = 1$ is $∣ - 1 - 1∣ = 2 = a$ ✓. Check an LR endpoint $(2, - 3)$ on the curve: $(x + 2)^{2} = (2 + 2)^{2} = 16$ and $- 8 (y + 1) = - 8 (- 3 + 1) = 16$ ✓.

The shifted down-parabola with its Quadrant-III focus and the two-signed latus rectum:

Recall

Recall Quick self-test — forecast each answer, then check

These questions each tie back to the "Match → Fourth → Place → Check" strategy; the answer explains why the point matters, not just what it is.

What does the letter $a$ mean on this page, and why keep it positive? ::: $a$ is the vertex-to-focus distance (equal to vertex-to-directrix distance); we keep it positive so it always means a size, letting the equation's sign carry direction. What is the axis of symmetry, and how do you spot it from the equation? ::: The line through the focus perpendicular to the directrix; if $y$ is squared the axis is horizontal, if $x$ is squared it is vertical. Right vs left opening — what actually flips, and why? ::: Only the sign of the coefficient flips; focus and directrix swap to opposite sides while $a$ stays positive — this is why we separate size ( $a$ ) from direction (the sign). How do you get $a$ from any coefficient, and why not use the raw number? ::: Divide the coefficient's magnitude by 4, because the form is written as $4 a$ so the latus rectum reads off directly — using the raw number is the single most common exam error. What is the focal-distance shortcut for a point on $y^{2} = 4 a x$ , and why is it legal? ::: It equals $x + a$ (the gap to the directrix $x = - a$ ); it's legal because the defining equality lets us swap the hard focus-distance for the easy line-distance. What happens geometrically as $a \to 0$ , and why does it stop being a parabola? ::: Focus meets the directrix and the curve flattens to the axis line — the definition forbids the directrix passing through the focus, so the object degenerates. For a shifted parabola, what is the one extra step and why? ::: Read the vertex $(h, k)$ from the shifted form and add $(h, k)$ to every feature — because sliding the origin moves focus, directrix and vertex rigidly together.

Connections

Parent: Parabola — standard forms (Hinglish)
Conic Sections — Overview
Ellipse — standard forms, Hyperbola — standard forms
Eccentricity and focus-directrix definition
Tangents and Normals to a Parabola
Coordinate Geometry — Distance Formula
Projectile Motion