MathsISC Class 11 · Conic Sections

The Parabola: Focus, Directrix, Latus Rectum and the Four Standard Forms

Updated 17 October 2026 · Fact-checked

A parabola is the set of points equidistant from a fixed point (focus) and a fixed line (directrix), with eccentricity e = 1. The standard form y² = 4ax has focus (a, 0), directrix x = −a, axis the x-axis, and latus rectum of length 4a. The other three forms are y² = −4ax, x² = 4ay and x² = −4ay.

Every point P on the curve is as far from the focus S as from the directrix (PS = PM); the latus rectum through S has length 4a.

Illustration drawn for QuestionBuddy.

Understand Parabola: Standard Forms and Equations

Definition. For a point P on the parabola, $P S = P M$ where S is the focus and M the foot of the perpendicular on the directrix: the eccentricity $e = 1$ . The line through the focus perpendicular to the directrix is the axis; the vertex is the point where the axis meets the curve, midway between the focus and the directrix.

Standard forms (vertex at the origin).

Equation	Axis	Focus	Directrix	Opens
$y^{2} = 4 a x$	x-axis	(a, 0)	$x = - a$	right
$y^{2} = - 4 a x$	x-axis	(−a, 0)	$x = a$	left
$x^{2} = 4 a y$	y-axis	(0, a)	$y = - a$	up
$x^{2} = - 4 a y$	y-axis	(0, −a)	$y = a$	down

Latus rectum. The chord through the focus perpendicular to the axis has length $4 a$ ; its end points for $y^{2} = 4 a x$ are $(a, \pm 2 a)$ .

Focal distance. For $P (x_{1}, y_{1})$ on $y^{2} = 4 a x$ , $S P = x_{1} + a$ . Parametric point: $(a t^{2}, 2 a t)$ .

Finding the equation. (1) Locate the focus and directrix, or the vertex and axis. (2) Decide the form from the direction in which it opens. (3) Find a from the focus (distance from the vertex) or from a point on the curve. If the data is the focus (p, q) and directrix, use $P S^{2} = P M^{2}$ directly.

Applications. The cross-section of a headlight reflector or satellite dish is a parabola; rays parallel to the axis meet at the focus.

Parabola: Standard Forms and Equations: formulas to remember

Standard form: $$y^2 = 4ax$: focus $(a, 0)$, directrix $x = -a$$; LR length $4 a$ .
Other forms: $$y^2 = -4ax$ | $x^2 = 4ay$ | $x^2 = -4ay$$; Direction of opening changes.
Focal distance: $$SP = x_1 + a$ for $y^2 = 4ax$$; Equal to the distance from the directrix.
Parametric form: $$(at^2, 2at)$$; For $y^{2} = 4 a x$ .

Parabola: Standard Forms and Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A satellite dish

Suppose a dish has a cross-section that is a parabola, 60 cm across the rim and 10 cm deep, with its vertex at the origin and axis along the x-axis.

Find its equation and the position of the receiver at the focus.

Try it first, then show the solutionSolution

The rim point is (10, 30), on $y^{2} = 4 a x$ : $900 = 4 a \times 10$ .
$a = 22.5$ cm, so the equation is $y^{2} = 90 x$ .
The receiver is at the focus (22.5, 0), 22.5 cm from the vertex along the axis.

Answer: y² = 90x; focus 22.5 cm from the vertex.

Memory hook: The focus lies on the axis at distance a from the vertex.

Tricks to remember Parabola: Standard Forms and Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Which variable is squared
y² gives a horizontal axis; x² gives a vertical axis.
Use it when: Identifying the form.
4a is the coefficient
Compare with 4a to get a, and use the sign for direction.
Use it when: Reading the focus and directrix.

Parabola: Standard Forms and Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Focus of a parabola · 1 mark · MCQBoard level

The focus of the parabola $y^{2} = 12 x$ is:
(a) (12, 0)
(b) (6, 0)
(c) (3, 0)
(d) (0, 3)

Try it first, then show the solutionSolution

$4 a = 12$ , so $a = 3$ and the focus is (3, 0).

Answer: (c) (3, 0)

Example 2: Elements of a parabola · 3 marks · Short answerBoard level

Find the focus, directrix and length of the latus rectum of $y^{2} = - 8 x$ .

Try it first, then show the solutionSolution

$4 a = 8$ , so $a = 2$ . (1 mark)
The parabola opens left: focus $(- 2, 0)$ and directrix $x = 2$ . (1 mark)
Length of the latus rectum $= 4 a = 8$ . (1 mark)

Answer: Focus (−2, 0); directrix x = 2; LR = 8.

Example 3: Equation from data · 4 marks · Short answerHarder board level

Find the equation of the parabola with focus (0, −3) and directrix y = 3, and the equation of the parabola with vertex at the origin, axis along the y-axis and passing through (6, −3).

Try it first, then show the solutionSolution

The vertex is midway at the origin, a = 3, and the parabola opens down: $x^{2} = - 12 y$ . (2 marks)
Axis along the y-axis and a point with negative y: $x^{2} = - 4 a y$ . (1 mark)
$36 = - 4 a (- 3) = 12 a$ , so $a = 3$ and $x^{2} = - 12 y$ . (1 mark)

Answer: x² = −12y (both cases).

Example 4: Focal distance · 4 marks · Short answerHarder board level

For the parabola $y^{2} = 16 x$ , find the point(s) with ordinate 8, the focal distance of the point, and the end points of the latus rectum.

Try it first, then show the solutionSolution

$a = 4$ . With $y = 8$ : $64 = 16 x$ , so $x = 4$ ; the point is (4, 8). (1.5 marks)
Focal distance $= x + a = 4 + 4 = 8$ . (1 mark)
Latus rectum end points $(a, \pm 2 a) = (4, \pm 8)$ , so (4, 8) is an end point and the length is $4 a = 16$ . (1.5 marks)

Answer: (4, 8); focal distance 8; LR end points (4, ±8).

Example 5: Reflector design · 5 marks · Long answerHOTS / competency

A parabolic reflector is 60 cm across at the opening and 10 cm deep. Find where the bulb should be placed, the length of the latus rectum and the height of the reflector section at a point 22.5 cm from the vertex along the axis.

Try it first, then show the solutionSolution

Take $y^{2} = 4 a x$ through (10, 30): $900 = 40 a$ , so $a = 22.5$ cm. (1.5 marks)
The bulb should be at the focus, 22.5 cm from the vertex along the axis. (1 mark)
Latus rectum $= 4 a = 90$ cm. (1 mark)
At $x = 22.5$ : $y^{2} = 90 \times 22.5 = 2025$ , $y = 45$ cm, so the width of the reflector at the focus is 90 cm (equal to the latus rectum). (1.5 marks)

Answer: Bulb at 22.5 cm; LR = 90 cm; half-height 45 cm at the focus.

Where students lose marks in Parabola: Standard Forms and Equations

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign for the direction
Writing the focus of y² = −8x as (2, 0).
Do this: The negative sign means the parabola opens left; the focus is (−2, 0).
Usually costs: 1 to 2 marks.
Mistake: Confusing a and 4a
Taking a = 12 for y² = 12x.
Do this: Compare with 4a: a = 3.
Usually costs: 2 marks.
Mistake: Using the wrong form
Using y² = 4ax for a vertical parabola.
Do this: Check which variable is squared.
Usually costs: 1 mark.

Exam tips

Sketch the parabola with the focus and the directrix.
Write 4a explicitly.
Check a point on the curve against your equation.
Mention the vertex, axis and focus in the final answer.

Practice questions on Parabola: Standard Forms and Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation of the directrix of $x^{2} = 8 y$ is:
(a) $y = 2$
(b) $y = - 2$
(c) $x = 2$
(d) $x = - 2$
Hint
4a = 8, so a = 2.
Answer
(b) $y = - 2$
Q2 · 2 marks · Short answerBoard level
Find the equation of the parabola with focus (3, 0) and vertex at the origin.
Hint
a = 3, opens right.
Answer
$y^{2} = 12 x$ .
Q3 · 3 marks · Short answerHarder board level
Find the equation of the parabola with vertex (0, 0), axis along x and passing through (2, −3).
Hint
The point has x > 0: opens right.
Answer
$y^{2} = 4 a x$ : $9 = 8 a$ , $a = \frac{9}{8}$ ; $y^{2} = \frac{9}{2} x$ .
Q4 · 3 marks · Short answerHarder board level
A point P on $y^{2} = 4 x$ has focal distance 5. Find P.
Hint
SP = x + a.
Answer
$x + 1 = 5$ , $x = 4$ , $y^{2} = 16$ , so $P = (4, \pm 4)$ .
Q5 · 3 marks · Short answerHarder board level
Find the vertex, focus and directrix of $(y - 2)^{2} = 8 (x + 1)$ .
Hint
Shift the origin.
Answer
Vertex $(- 1, 2)$ ; $a = 2$ so focus $(1, 2)$ ; directrix $x = - 3$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a bridge arch is a parabola $x^{2} = - 20 y$ (metres), with the highest point at the origin and the road at ground level y = −5.
(i) Find the span at the road level. (1 mark)
(ii) Find the focus. (1 mark)
(iii) Find the height of the arch above the road at a horizontal distance of 5 m from the centre. (1 mark)
(iv) Find the directrix. (1 mark)
Hint
Put y = −5 for the span.
Answer
(i) $x^{2} = 100$ , $x = \pm 10$ : span 20 m. (ii) $a = 5$ : focus (0, −5). (iii) $x = 5$ : $y = - \frac{25}{20} = - 1.25$ ; height above the road $= 5 - 1.25 = 3.75$ m. (iv) $y = 5$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Length of the latus rectum?Show answer
4a.
2.Eccentricity of a parabola?Show answer
1.
3.Focus of x² = 4ay?Show answer
(0, a).
4.Directrix of y² = 4ax?Show answer
x = −a.

Parabola: Standard Forms and Equations: frequently asked questions

Why does a dish need a parabola?

All rays parallel to the axis reflect to the focus, so signals concentrate at the receiver.

What if the vertex is not at the origin?

Shift the origin: for vertex (h, k), use (y - k)^2 = 4a(x - h).

What is a double ordinate?

A chord perpendicular to the axis.