Topic drillAbout 15 minutesTested in all 5 recent real CSCA papers (8 questions)

Focus and Directrix of a Parabola

The focus and the directrix sit on opposite sides of the vertex, both p/2 away. See the four standard forms on a graph, then learn the routine: put it in standard form, read p, write the focus and directrix, and turn a distance to the focus into a distance to the directrix. With CSCA exam examples, the two most common mistakes and a 5-question mastery check.

This page is written in Chinese and English (the two CSCA exam languages); you are reading the English version.

1Warm-up

Start with a real exam question, on instinct:

Real examSource: CSCA Math December 2025 Past Paper

Which statement about the parabola

y^2=4x

is correct?

2The idea: focus and directrix on opposite sides of the vertex, equally far

Every point of a parabola is as far from the focus as from the directrix. There are just four standard forms, differing only in the opening: $y^2=2px$ opens right with focus $\left(\frac p2,0\right)$ and directrix $x=-\frac p2$ ; the others are the same picture turned.

Pick the opening, then drag $p$ and P: watch the focus, directrix and axis move, and P's two distances stay equal.

Heavy line: directrix. Dashed: axis. F: focus. Small dot: vertex. Solid segment joins P to the focus, dashed one joins P to the directrix.

p = 2P: 3

$x^2=4y\quad(y=\frac{1}{4}x^2)$
$2p=4,\ p=2,\ \dfrac p2=1$ : focus and directrix are both $1$ from the vertex
focus $\left(0,1\right)$ , directrix $y=-1$ , axis the y-axis
$|PF|=4$ , distance to the directrix $=4$ : equal

Three steps to the focus and directrix

standard form $y^2=\pm2px$ or $x^2=\pm2py$

first-power coefficient $=2p$ : find $p$

focus and directrix both $\frac p2$ from the vertex, on opposite sides

$p$ , $2p$ and $\frac p2$ are three different numbers, and the easiest to mix up.

The four openings

Opens	Equation	Focus	Directrix
right	$y^2=2px$	$\left(\frac p2,0\right)$	$x=-\frac p2$
left	$y^2=-2px$	$\left(-\frac p2,0\right)$	$x=\frac p2$
up	$x^2=2py$	$\left(0,\frac p2\right)$	$y=-\frac p2$
down	$x^2=-2py$	$\left(0,-\frac p2\right)$	$y=\frac p2$

The signs of the focus and the directrix are opposite; the vertex is always the origin.

Two conventions: do not be thrown by a different $p$ in a solution

Convention	Equation	What $p$ is	For $y^2=4x$
This page (common in textbooks)	$y^2=2px$	focus-to-directrix distance	$p=2$ , focus $\left(\frac p2,0\right)=(1,0)$
Some solutions	$y^2=4px$	vertex-to-focus distance	$p=1$ , focus $(p,0)=(1,0)$

If an original solution says $y^2=4px$ with $p=1$ , nothing is wrong: its $p$ means something different (it is this page's $\frac p2$ ). The focus and directrix come out the same, so the answer is unaffected.

Distance questions: swap "to the focus" for "to the directrix"

$P$ on $y^2=2px$

$|PF|=$ the distance from $P$ to the directrix

$=x_0+\frac p2$

$A$ inside the parabola (the focus side)

$|PA|+|PF|=|PA|+$ distance to the directrix $\ge$ the distance from $A$ to the directrix

minimum $=$ the distance from $A$ to the directrix, with $P$ on the perpendicular through $A$

$A$ outside the parabola (on the directrix counts)

$|PA|+|PF|\ge|AF|$

minimum $=|AF|$ , with $P$ where segment $AF$ meets the parabola

Which bound to use depends only on where $A$ is, not on whether the problem says $|PF|$ or "distance to the directrix". To decide, put $A$ into $y^2-2px$ : below $0$ is inside, above $0$ is outside.

Read it as text

How to read it: the squared letter tells you the axis. $y^2=\dots$ opens along the $x$ -axis (left or right); $x^2=\dots$ opens along the $y$ -axis (up or down). The sign of the first-power term gives the direction and its coefficient is $2p$ .

Exams often write $y=\frac14x^2$ : first put it in standard form $x^2=4y$ , then read $2p=4$ , $p=2$ , focus $(0,1)$ , directrix $y=-1$ . Mind it: $2p$ is the coefficient, and the focus is $\frac p2$ from the vertex, not $p$ and not $2p$ .

For "the distance from a point to the focus", switch to "the distance to the directrix": for $y^2=2px$ , $|PF|=x_0+\frac p2$ . To minimise $|PA|+|PF|$ — which is the same quantity as $|PA|$ plus the distance from $P$ to the directrix — there are two lower bounds: $|PA|+|PF|\ge|AF|$ and $|PA|+d(P,\text{directrix})\ge d(A,\text{directrix})$ . The minimum is the one that can be reached, and only the position of $A$ decides which, however the problem is worded: if $A$ is inside the parabola (on the focus side) it is the distance from $A$ to the directrix, with $P$ where the line through $A$ perpendicular to the directrix meets the parabola; if $A$ is outside (a point on the directrix counts) it is $|AF|$ , with $P$ where segment $AF$ meets the parabola.

34 steps

Walking through $y=\dfrac14x^2$

1
Put it in standard form
$y=\dfrac14x^2\ \Rightarrow\ x^2=4y$
- $x^2$ on the left → opens along the $y$ -axis
- $+4y$ → up
2
Read $2p$ and $p$
$2p=4\ \Rightarrow\ p=2,\quad \dfrac p2=1$
The coefficient is $2p$ , not $p$ ; the distance from the vertex is $\frac p2$ .
3
Write the focus and directrix
- focus $(0,1)$ ( $1$ above the vertex)
- directrix $y=-1$ ( $1$ below the vertex)
The directrix is perpendicular to the axis: the axis is the $y$ -axis, so the directrix is a horizontal line $y=\dots$ .
4
For distances, switch to the directrix
- $|PF|=$ distance to the directrix
- for the distance from the focus to a point, write the focus first, then use the distance formula
- min of $|PA|+|PF|$ : look at where $A$ is. Inside → the distance to the directrix; outside → $|AF|$

Read it as text

1. Put it in standard form
Rewrite it as $y^2=\pm2px$ or $x^2=\pm2py$ : the squared letter fixes the axis.
2. Read $2p$ and $p$
The first-power coefficient is $2p$ ; then find $p$ and $\frac p2$ .
3. Write the focus and directrix
The focus is $\frac p2$ from the vertex on the opening side, the directrix $\frac p2$ on the other, perpendicular to the axis.
4. For distances, switch to the directrix
Replace $|PF|$ by the distance to the directrix; for the distance from the focus to a point, write the focus and use the distance formula. The minimum of $|PA|+|PF|$ depends on where $A$ is: inside, the distance from $A$ to the directrix; outside, $|AF|$ , however the problem is worded.

Choose first, then check: 5 parabolas

Commit to each step first; a wrong pick tells you why.

Step 1Step 1, standard form:

Read all 5 worked examples in full

$y=\frac14x^2$

Step 1Step 1, standard form:
$x^2=4y$ : opens up.
✗ $y^2=4x$ :The squared letter cannot swap: in $y=\frac14x^2$ it is $x$ that is squared, so $x^2$ stays on the left.
✗ $x^2=\dfrac14y$ :Multiply both sides by $4$ : $4y=x^2$ , not divide.
Step 2Step 2, read $p$ :
$2p=4$ , $p=2$ , $\frac p2=1$ .
✗ $p=4$ : $4$ is $2p$ , not $p$ . The standard form is $x^2=2py$ .
✗ $p=\dfrac12$ : $p=\frac12$ would be $x^2=y$ . Here the coefficient is $4$ .
Step 3Step 3, the focus:
Focus $(0,\frac p2)=(0,1)$ .
✗ $(1,0)$ :The axis of $x^2=4y$ is the $y$ -axis, so the focus is on it: $(0,\cdot)$ .
✗ $(0,2)$ :The focus is $\frac p2=1$ from the vertex, not $p=2$ .
Step 4Step 4, the directrix:
Directrix $y=-\frac p2=-1$ .
✗ $y=1$ :The directrix and focus are on opposite sides of the vertex: focus above, so directrix below, $y=-1$ .
✗ $x=-1$ :The directrix is perpendicular to the axis: the axis is the $y$ -axis, so the directrix is horizontal, $y=\dots$ , not $x=\dots$ .

Result: $F(0,1),\quad y=-1$

$y^2=-8x$

Step 1Step 1, the opening:
$y^2=-8x$ opens left.
✗ down: $y^2=\dots$ opens along the $x$ -axis (left or right); the minus sign says left. Down would be $x^2=-\dots y$ .
Step 2Step 2, read $p$ :
$\frac p2=2$ .
✗ $p=8$ : $8$ is $2p$ . $p=4$ .
Step 3Step 3, the directrix:
Directrix $x=\frac p2=2$ .
✗ $x=-2$ :The focus is $(-2,0)$ , on the left; the directrix is on the other side, $x=2$ .
✗ $x=-4$ :It is $\frac p2=2$ from the vertex, not $p=4$ .

Result: $x=2$

$x=\frac18y^2$

Step 1Step 1, standard form:
$y^2=8x$ , opens right.
✗ $y^2=\dfrac18x$ :Multiply both sides by $8$ : $8x=y^2$ .
✗ $x^2=8y$ :It is $y$ that is squared here, not $x$ .
Step 2Step 2, the directrix:
Directrix $x=-\frac p2=-2$ .
✗ $x=-4$ : $2p=8$ , $p=4$ , and the distance from the vertex is $\frac p2=2$ , not $4$ .
✗ $y=-2$ :The axis is the $x$ -axis, so the directrix is perpendicular to it: a vertical line $x=\dots$ .

Result: $x=-2$

$y^2=4x,\ x_P=4$

Step 1Step 1, the directrix:
Directrix $x=-1$ .
✗ $x=-2$ : $2p=4$ , $p=2$ , directrix $x=-\frac p2=-1$ .
Step 2Step 2, $|PF|=$ the distance from $P$ to the directrix:
$|PF|=x_0+\frac p2=5$ .
✗ $4$ :That is the distance to the $y$ -axis. To the directrix $x=-1$ you add $\frac p2=1$ .
✗ $3$ : $4-1$ is the gap to the focus's $x$ -coordinate. The directrix is at $x=-1$ , so the distance is $4-(-1)=5$ .

Result: $|PF|=5$

$x^2=8y,\ F\to(2,5)$

Step 1Step 1, the focus:
Focus $(0,\frac p2)=(0,2)$ .
✗ $(2,0)$ :That is the focus of $y^2=8x$ . The axis of $x^2=8y$ is the $y$ -axis, so the focus is on it.
✗ $(0,4)$ : $2p=8$ , $p=4$ , and the focus is $\frac p2=2$ up, not $p=4$ .
Step 2Step 2, the distance to $(2,5)$ :
$\sqrt{(2-0)^2+(5-2)^2}=\sqrt{13}$ .
✗ $3$ : $5-2=3$ is only the vertical gap; the horizontal gap $2$ counts too: use the distance formula.
✗ $\sqrt{29}$ : $\sqrt{2^2+5^2}$ is the distance to the origin (the vertex), not to the focus.

Result: $\sqrt{13}$

4Worked examples

Try it yourself, then open the steps

Real examSource: CSCA Math March 2026 Past Paper

The equation of the directrix of the parabola

y = - x ^ { 2 } / 4

Step 1Standard form first: $y=-\dfrac{x^2}{4}$ is $x^2=-4y$ , opening down.
Step 2 $2p=4$ , $p=2$ , $\dfrac p2=1$ : the focus is $1$ below the vertex, $(0,-1)$ .
Step 3The directrix is on the other side, $1$ above the vertex: $y=1$ . Answer C.

Try it yourself, then open the steps

MockSource: Math Mock Exam 13

Find the directrix of

x=4y^2

Step 1 $x=4y^2$ becomes $y^2=\dfrac14x$ : $y$ is squared, so it opens along the $x$ -axis, to the right.
Step 2 $2p=\dfrac14$ , $p=\dfrac18$ , $\dfrac p2=\dfrac1{16}$ .
Step 3The focus is on the right, so the directrix is on the left: $x=-\dfrac1{16}$ . Answer D.

Try it yourself, then open the steps

Real examSource: CSCA Math January 2026 Past Paper

Let the focus of the parabola

y ^ { 2 } = 4 x

F

. Suppose that a point

P

is on the parabola, and the horizontal coordinate of

P

is 4. Then

| P F | = ( )

Step 1 $y^2=4x$ : $2p=4$ , $p=2$ , directrix $x=-1$ .
Step 2On a parabola, the distance to the focus equals the distance to the directrix: $|PF|=4-(-1)=5$ .
Step 3Or with the formula $|PF|=x_0+\dfrac p2=4+1=5$ . Answer D.

5Common mistakes

Answer data: CSCA.app students' answers over the 90 days before 2026-10-23.

1. Mixing up where the focus, directrix, axis and vertex are

What the data shows:On the June 2026 real exam, asked which statement about $y=\frac{x^2}{4}$ is correct, only about a third get it right (focus $(0,1)$ ). About a quarter choose "the axis is the $x$ -axis", about a quarter choose "the directrix is $y=1$ ", and about one in eight choose "the vertex is $(1,0)$ ".

How to avoid it:Put it in standard form $x^2=4y$ and all four are settled at once: the vertex is the origin; $x^2$ is on the left, so it opens along the $y$ -axis and the axis is the $y$ -axis; $2p=4$ , $p=2$ , the focus is $\frac p2=1$ above the vertex, $(0,1)$ ; the directrix is on the other side, $y=-1$ . Focus and directrix sit on opposite sides, with opposite signs.

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

About the parabola

y = \frac{x^2}{4}

, which of the following statements is correct?

2. Finding the wrong focus, or leaving out a direction in the distance

What the data shows:On a mock question, the distance from the focus of $x^2=8y$ to the point $(2,5)$ : fewer than half get it right ( $\sqrt{13}$ ). About one in five choose $\sqrt{29}$ , exactly the distance from $(2,5)$ to the origin; about one in six choose $5$ , the distance from $(2,5)$ to $(2,0)$ , and $(2,0)$ is the focus of $y^2=8x$ ; about one in seven choose $3$ , only the vertical gap.

How to avoid it:Write the focus first: $x^2=8y$ , $2p=8$ , $p=4$ , the focus is on the $y$ -axis, $\frac p2=2$ above the vertex: $(0,2)$ . Then use the distance formula: $\sqrt{(2-0)^2+(5-2)^2}=\sqrt{13}$ ; both coordinate gaps count. $x^2$ on the left means the focus is on the $y$ -axis; do not confuse it with the focus $(2,0)$ of $y^2=8x$ .

Check yourself

MockSource: Math Mock Exam 1

Find the distance from the focus of the parabola

x^2=8y

to the point

(2,5)

6Practice

Question 1 of 5

MockSource: Math Mock Exam 15

Which parabola opens to the left?

0 of 5 right in this set

7Mastery check

5 questions; 4 right means you have it. No solutions until the end.

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Quadratic Functions and Equations

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Conic Sections: Basic Quantities

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Chapter practice

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1Warm-up

2The idea: focus and directrix on opposite sides of the vertex, equally far

Three steps to the focus and directrix

The four openings

Two conventions: do not be thrown by a different ppp in a solution

Distance questions: swap "to the focus" for "to the directrix"

34 steps

Walking through y=14x2y=\dfrac14x^2y=41​x2

Choose first, then check: 5 parabolas

y=14x2y=\frac14x^2y=41​x2

y2=−8xy^2=-8xy2=−8x

x=18y2x=\frac18y^2x=81​y2

y2=4x, xP=4y^2=4x,\ x_P=4y2=4x, xP​=4

x2=8y, F→(2,5)x^2=8y,\ F\to(2,5)x2=8y, F→(2,5)

4Worked examples

5Common mistakes

1. Mixing up where the focus, directrix, axis and vertex are

2. Finding the wrong focus, or leaving out a direction in the distance

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

Two conventions: do not be thrown by a different $p$ in a solution

Walking through $y=\dfrac14x^2$

$y=\frac14x^2$

$y^2=-8x$

$x=\frac18y^2$

$y^2=4x,\ x_P=4$

$x^2=8y,\ F\to(2,5)$