MathsCBSE Class 11 · Conic Sections

Parabola: Focus, Directrix and Latus Rectum (Class 11 Conic Sections)

Q: How do I know which way a parabola opens?

If y is squared, it opens right for a positive coefficient of x and left for a negative one. If x is squared, it opens up for a positive coefficient of y and down for a negative one.

Q: Is the vertex always at the origin?

In the standard forms studied in Class 11, yes. Shifting the origin is not part of the syllabus.

Q: What is the eccentricity of a parabola?

It is 1, because the distance from the focus equals the distance from the directrix.

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 vertex at the origin it has four standard forms: y² = 4ax, y² = −4ax, x² = 4ay and x² = −4ay (a > 0). The focus is at distance a from the vertex, the directrix is the line at distance a on the other side, and the latus rectum has length 4a. The focal distance of (x1, y1) on y² = 4ax is x1 + a.

Understand Parabola: Focus, Directrix and Latus Rectum

A parabola is the locus of a point that moves so that its distance from a fixed point $F$ (the focus) equals its distance from a fixed line $l$ (the directrix), where $F$ is not on $l$ . Its axis is the line through $F$ perpendicular to $l$ , and the vertex is the point of the curve on the axis, midway between $F$ and $l$ .

Standard forms (vertex at the origin, $a > 0$ ):

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

To read a parabola from its equation, match it with $4 a$ : the coefficient of the linear term is $\pm 4 a$ . For $y^{2} = - 12 x$ , $4 a = 12$ , so $a = 3$ .

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

Focal distance. For a point $(x_{1}, y_{1})$ on $y^{2} = 4 a x$ , the distance from the focus equals the distance from the directrix: $x_{1} + a$ . Similarly, for $x^{2} = 4 a y$ it is $y_{1} + a$ .

Finding the equation. Decide the orientation from the data (where is the focus or the point?), write the standard form with unknown $a$ , and use the given condition (focus, directrix, point on the curve, latus rectum length) to find $a$ . If the question gives a point such as $(6, - 3)$ with the axis along the $y$ -axis, the parabola must open downward, so use $x^{2} = - 4 a y$ .

Locus from the definition. If asked for the locus of a point equidistant from $(0, 3)$ and the line $y = - 3$ , set $x^{2} + (y - 3)^{2} = ∣ y + 3∣$ , square, and simplify to $x^{2} = 12 y$ .

Applications. Parabolic mirrors place the source at the focus so that rays leave parallel to the axis, and arches, cables and reflectors are modelled with the vertex at the origin. The latus rectum is the width of the parabola at the level of the focus.

Parabola: Focus, Directrix and Latus Rectum: formulas to remember

Parabola opening right: $y^{2} = 4 a x; f oc u s (a, 0); d i r ec t r i xx = - a$; Latus rectum 4a.
Parabola opening up: $x^{2} = 4 a y; f oc u s (0, a); d i r ec t r i x y = - a$; Replace y by −y or x by −x for the others.
Latus rectum: $L e n g t h 4 a; e n d s (a, \pm 2 a) f or y^{2} = 4 a x$; Chord through the focus perpendicular to the axis.
Focal distance: $x 1 + a f or a p o in t (x 1, y 1) o n y^{2} = 4 a x$; Equals the distance to the directrix.
Locus definition: $D i s t an ce t o f oc u s = d i s t an ce t o d i r ec t r i x$; Square both sides to get the equation.

How to solve Parabola: Focus, Directrix and Latus Rectum questions

Use these steps for reading a parabola and for finding its equation.

1Rewrite the equation as one of the four standard forms and compare the coefficient with $\pm 4 a$ to find $a$ .
2Decide the direction of opening from the sign and from which variable is squared.
3Write the focus, directrix and length of the latus rectum from the table.
4For an equation, choose the form that fits the orientation of the given focus, directrix or point, and substitute the data to get $a$ .
5For a focal distance, use $x_{1} + a$ (or $y_{1} + a$ ) instead of the distance formula.
6Check that the point and focus you found satisfy the definition.

Parabola: Focus, Directrix and Latus Rectum 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.

Parabolic reflector

Suppose a torch has a reflector shaped like a parabola, with its vertex at the origin and the axis along the positive $x$ -axis. The reflector is 4 cm deep and its rim is 12 cm across, so a rim point is $(4, 6)$ .

Find the equation of the reflector and the distance of the bulb (placed at the focus) from the vertex.

Try it first, then show the solutionSolution

Take $y^{2} = 4 a x$ through $(4, 6)$ : $36 = 16 a$ , so $a = \frac{9}{4} = 2.25$ .
Equation: $y^{2} = 9 x$ .
The bulb is at the focus $(2.25, 0)$ , that is, 2.25 cm from the vertex.
At the level of the focus the reflector is $4 a = 9$ cm wide (the latus rectum).

Answer: $y^{2} = 9 x$ ; bulb 2.25 cm from the vertex.

Memory hook: Rim point gives a, and a is the focus distance.

Locus from the definition

Suppose a field is watered by a sprinkler at $(0, 3)$ and a straight canal along $y = - 3$ . A flag is to be placed at every point that is exactly as far from the sprinkler as from the canal.

Find the equation of the curve on which the flags lie.

Try it first, then show the solutionSolution

Let a flag be at $(x, y)$ . Distance to sprinkler $= x^{2} + (y - 3)^{2}$ , distance to canal $= ∣ y + 3∣$ .
Equate and square: $x^{2} + y^{2} - 6 y + 9 = y^{2} + 6 y + 9$ .
$x^{2} = 12 y$ , a parabola with $a = 3$ .

Answer: $x^{2} = 12 y$

Memory hook: Equal distance to a point and a line always means parabola.

Tricks to remember Parabola: Focus, Directrix and Latus Rectum

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

Read a from the coefficient
The number in front of the linear term is ±4a; divide by 4.
Use it when: Every standard-form question.
The other letter is the axis
If y is squared, the axis is the x-axis; if x is squared, the axis is the y-axis.
Use it when: Picking the right form.
Focus + a on the far side
Directrix is the mirror of the focus about the vertex.
Use it when: Writing the directrix quickly.
Focal distance shortcut
Skip the distance formula: focal distance = coordinate + a (with the right sign).
Use it when: Points on a parabola at a given focal distance.

Parabola: Focus, Directrix and Latus Rectum: solved examples

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

Example 1: Reading a parabola · 2 marks · Very short answerBoard level

Find the focus, directrix, axis and length of the latus rectum of $5 x^{2} + 30 y = 0$ .

Try it first, then show the solutionSolution

Rewrite as $x^{2} = - 6 y$ , which has the form $x^{2} = - 4 a y$ with $4 a = 6$ , so $a = \frac{3}{2}$ . (½ mark)
Focus $(0, - \frac{3}{2})$ ; directrix $y = \frac{3}{2}$ . (1 mark)
Axis: the $y$ -axis; latus rectum $= 4 a = 6$ . (½ mark)

Answer: Focus $(0, - \frac{3}{2})$ ; directrix $y = \frac{3}{2}$ ; axis the $y$ -axis; LR $= 6$

Example 2: Parabola from the latus rectum · 3 marks · Short answerBoard level

A parabola has its vertex at the origin, is symmetric about the $x$ -axis, opens to the left and has a latus rectum of length 20. Find its equation, focus and directrix, and the ends of the latus rectum.

Try it first, then show the solutionSolution

Form: $y^{2} = - 4 a x$ with $4 a = 20$ , so $a = 5$ . (1 mark)
Equation $y^{2} = - 20 x$ ; focus $(- 5, 0)$ ; directrix $x = 5$ . (1 mark)
Ends of the latus rectum: $x = - 5$ , $y^{2} = 100$ , so $(- 5, \pm 10)$ . (1 mark)

Answer: $y^{2} = - 20 x$ ; focus $(- 5, 0)$ ; directrix $x = 5$ ; LR ends $(- 5, \pm 10)$

Example 3: Focal distance · 3 marks · Short answerHarder board level

A point $P$ on $y^{2} = 16 x$ is 13 units from the focus. Find the coordinates of $P$ .

Try it first, then show the solutionSolution

$4 a = 16$ , so $a = 4$ ; focus $(4, 0)$ . (½ mark)
Focal distance $= x + a = x + 4 = 13$ , so $x = 9$ . (1 mark)
$y^{2} = 16 \times 9 = 144$ , so $y = \pm 12$ . (1 mark)
Check: distance from $(4, 0)$ to $(9, 12)$ is $25 + 144 = 13$ . (½ mark)

Answer: $P = (9, 12)$ or $(9, - 12)$

Example 4: Locus from the definition · 3 marks · Short answerHarder board level

Find the locus of a point that is equidistant from the point $(0, 3)$ and the line $y = - 3$ .

Try it first, then show the solutionSolution

Let the point be $(x, y)$ . Distance to $(0, 3)$ : $x^{2} + (y - 3)^{2}$ ; distance to the line: $∣ y + 3∣$ . (1 mark)
Square both sides: $x^{2} + y^{2} - 6 y + 9 = y^{2} + 6 y + 9$ . (1 mark)
Simplify: $x^{2} = 12 y$ , a parabola with focus $(0, 3)$ and directrix $y = - 3$ . (1 mark)

Answer: $x^{2} = 12 y$

Example 5: Case study: a parabolic arch · 4 marks · Case-basedHOTS / competency

Suppose a bridge arch is a parabola with its highest point at the origin and its axis vertical (opening downward). The arch is 20 m wide at the base and the top is 8 m above the base, so the base ends are $(\pm 10, - 8)$ .
(i) Find the equation of the arch. (1 mark)
(ii) Find the height of the arch above the base at 4 m from the centre line. (1 mark)
(iii) Find the width of the arch at a height of 5 m above the base. (1 mark)
(iv) Find the position of the focus. (1 mark)

Try it first, then show the solutionSolution

(i) $x^{2} = - 4 a y$ through $(10, - 8)$ : $100 = 32 a$ , $a = \frac{25}{8}$ , so $x^{2} = - 12.5 y$ . (1 mark)
(ii) $x = 4$ : $y = - \frac{16}{12.5} = - 1.28$ ; height above base $= 8 - 1.28 = 6.72$ m. (1 mark)
(iii) 5 m above the base is $y = - 3$ : $x^{2} = 37.5$ , $x = \pm 6.124$ , so the width is about $12.25$ m. (1 mark)
(iv) Focus at $(0, - \frac{25}{8}) = (0, - 3.125)$ : 3.125 m below the top. (1 mark)

Answer: (i) $x^{2} = - 12.5 y$ (ii) 6.72 m (iii) about 12.25 m (iv) 3.125 m below the top

Where students lose marks in Parabola: Focus, Directrix and Latus Rectum

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

Mistake: Using the wrong form for the direction
The sign of the given point is ignored.
Wrong working
Point $(6, - 3)$ used in $x^{2} = 4 a y$ , giving $a = 3$ and $x^{2} = 12 y$ .
Right working
Negative $y$ means opening downward: $x^{2} = - 4 a y$ , so $x^{2} = - 12 y$ .
Do this: Sketch the direction first.
Usually costs: The whole equation.
Mistake: Taking 4a as a
The coefficient of the linear term is 4a.
Wrong working
$y^{2} = 12 x$ gives focus $(12, 0)$ .
Right working
$4 a = 12$ , $a = 3$ , focus $(3, 0)$ .
Do this: Write " $4 a = \dots$ " on a separate line.
Usually costs: Focus, directrix and LR.
Mistake: Wrong sign on the directrix
The directrix is on the opposite side of the vertex from the focus.
Wrong working
$y^{2} = - 12 x$ : directrix $x = - 3$ .
Right working
The focus is $(- 3, 0)$ , so the directrix is $x = 3$ .
Do this: Directrix is the mirror of the focus in the vertex.
Usually costs: 1 mark.
Mistake: Forgetting to reject impossible points
A root of the algebra may not lie on the curve.
Wrong working
Keeping $x = - 9$ on $y^{2} = 12 x$ .
Right working
$y^{2} = - 108 < 0$ , so no real point.
Do this: Substitute back into the curve.
Usually costs: The last mark.

Exam tips

Always compute $a$ first and write it down; every other quantity follows from it.
Sketch a small diagram with the focus, directrix and vertex for equation-building questions.
Use focal distance $= x_{1} + a$ rather than the distance formula; it saves time.
In locus questions finish by naming the curve and giving its focus and directrix.
Conic sections carry a good share of Unit III of the annual paper, so practise one equation-building question for every standard form.

Practice questions on Parabola: Focus, Directrix and Latus Rectum

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 focus of the parabola $x^{2} = - 8 y$ is:
(a) $(0, - 2)$
(b) $(0, 2)$
(c) $(- 2, 0)$
(d) $(2, 0)$
Hint
$4 a = 8$ and the parabola opens downward.
Answer
(a). $a = 2$ , focus $(0, - 2)$ .
Q2 · 1 mark · MCQHarder board level
The ends of the latus rectum of the parabola $x^{2} = - 24 y$ are:
(a) $(\pm 6, - 6)$
(b) $(\pm 12, - 6)$
(c) $(\pm 12, 6)$
(d) $(\pm 24, - 6)$
Hint
Here $4 a = 24$ and the LR lies on the line $y = - a$ .
Answer
(b). $a = 6$ ; the LR is on $y = - 6$ with $x^{2} = 144$ , so the ends are $(\pm 12, - 6)$ .
Q3 · 3 marks · Short answerHarder board level
Find the points on $y^{2} = 8 x$ whose focal distance is 6.
Hint
Here $a = 2$ and focal distance $= x + 2$ .
Answer
$x + 2 = 6$ gives $x = 4$ ; $y^{2} = 32$ , $y = \pm 42$ . Points $(4, \pm 42)$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For the parabola $x^{2} = 20 y$ the focus is 5 units from the vertex and the latus rectum has length 20.
Reason (R): For $x^{2} = 4 a y$ the focus is $(0, a)$ and the latus rectum has length $4 a$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find $a$ from $4 a = 20$ .
Answer
(a). $a = 5$ , so the focus is $(0, 5)$ and the latus rectum is $4 a = 20$ .
Q5 · 4 marks · Short answerHOTS / competency
A circle has its centre at the focus of $y^{2} = 12 x$ and passes through the ends of the latus rectum. Find its equation and the points where it meets the parabola.
Hint
Ends of the latus rectum of $y^{2} = 4 a x$ are $(a, \pm 2 a)$ .
Answer
Focus $(3, 0)$ ; ends of LR $(3, \pm 6)$ ; radius $6$ . Circle $(x - 3)^{2} + y^{2} = 36$ . Substituting $y^{2} = 12 x$ : $x^{2} + 6 x - 27 = 0$ , so $x = 3$ or $x = - 9$ . $x = - 9$ gives $y^{2} < 0$ , rejected. The curves meet only at $(3, \pm 6)$ .

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.What is the focus of y² = 4ax?Show answer
(a, 0).
2.What is the length of the latus rectum?Show answer
4a.
3.What is the directrix of x² = 4ay?Show answer
y = −a.
4.What is the focal distance of (x1, y1) on y² = 4ax?Show answer
x1 + a.
5.Where is the vertex relative to the focus and directrix?Show answer
Exactly midway between them.

Parabola: Focus, Directrix and Latus Rectum: frequently asked questions

How do I know which way a parabola opens?

If y is squared, it opens right for a positive coefficient of x and left for a negative one. If x is squared, it opens up for a positive coefficient of y and down for a negative one.

What is the latus rectum?

The chord through the focus perpendicular to the axis. Its length is 4a.

Is the vertex always at the origin?

In the standard forms studied in Class 11, yes. Shifting the origin is not part of the syllabus.

What is the eccentricity of a parabola?

It is 1, because the distance from the focus equals the distance from the directrix.