1. What is a parabola?

A parabola is the set of points that are equally distant from:

• A fixed point called the focus.
• A fixed line called the directrix.

The focus does not lie on the directrix.

For any point P on a parabola:

\[ PF=d(P,\text{directrix}). \]

The distance from a point to a line is the perpendicular distance.

Understanding the definition

If a point is 5 units from the focus and 5 units from the directrix, it lies on the parabola.

If those distances are different, it does not lie on the parabola.

Unlike an ellipse or hyperbola, a parabola has one focus.

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2. The main parts of a parabola

Vertex

The vertex is the turning point of the parabola.

It lies halfway between the focus and the directrix, measured along the axis of symmetry.

Focus

The focus is the fixed point used in the distance definition.

It lies inside the opening of the parabola.

Directrix

The directrix is the fixed line used in the definition.

It lies on the opposite side of the vertex from the focus.

Axis of symmetry

The axis passes through the vertex and focus and is perpendicular to the directrix.

The two sides of the parabola are mirror images across this line.

Opening direction

A parabola opens:

• Right or left when its axis is horizontal.
• Upward or downward when its axis is vertical.

It opens from the vertex toward the focus.

The parameter p

This lesson uses the standard forms containing 4p.

The parameter p is a signed quantity:

• Its magnitude, |p|, is the vertex-to-focus distance.
• Its sign determines the opening direction.

For a nondegenerate parabola:

\[ p\neq0. \]

The distance between the focus and directrix is:

\[ 2|p|. \]

Notation warning: Some textbooks use equations containing 2p instead of 4p. Always identify the convention before reading the focus coordinates.

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3. Deriving a standard equation

Consider a parabola with:

• Focus F = (p, 0).
• Directrix x = −p.
• Vertex at the origin.

Let P = (x, y) be a point on the parabola.

Step 1: Write the distance to the focus

\[ PF=\sqrt{(x-p)^2+y^2}. \]

Step 2: Write the distance to the directrix

The perpendicular distance to x = −p is:

\[ |x+p|. \]

Step 3: Set the distances equal

\[ \sqrt{(x-p)^2+y^2}=|x+p|. \]

Both sides are nonnegative, so squaring preserves the equality:

\[ (x-p)^2+y^2=(x+p)^2. \]

Expand:

\[ x^2-2px+p^2+y^2=x^2+2px+p^2. \]

Simplify:

\[ y^2=4px. \]

This is the standard equation of a horizontal parabola with vertex at the origin.

Interchanging x and y gives the vertical form:

\[ x^2=4py. \]

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4. Standard equations centered at the origin

4.1 Horizontal axis

\[ y^2=4px. \]

Its properties are:

• Vertex: (0, 0).
• Focus: (p, 0).
• Directrix: x = −p.
• Axis of symmetry: y = 0.

If p > 0, it opens right.

If p < 0, it opens left.

4.2 Vertical axis

\[ x^2=4py. \]

Its properties are:

• Vertex: (0, 0).
• Focus: (0, p).
• Directrix: y = −p.
• Axis of symmetry: x = 0.

If p > 0, it opens upward.

If p < 0, it opens downward.

Standard-form comparison

8. Worked example 3: Reading a translated equation

Find the main properties of:

\[ (y+2)^2=-12(x-1). \]

Step 1: Identify the vertex

Compare with:

\[ (y-k)^2=4p(x-h). \]

Therefore:

\[ h=1,\qquad k=-2. \]

Vertex: (1, −2).

Step 2: Find p

\[ 4p=-12, \qquad p=-3. \]

The parabola opens left.

Step 3: Find the focus

\[ (h+p,k)=(1-3,-2). \]

Focus: (−2, −2).

Step 4: Find the directrix

\[ x=h-p=1-(-3). \]

Directrix: x = 4.

Axis of symmetry: y = −2.

Verification

The vertex is 3 units from the focus and 3 units from the directrix.

The focus lies to the left of the vertex, consistent with the opening direction.

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9. Constructing a parabola equation

Method A: Given the vertex and focus

1. Determine whether the axis is horizontal or vertical.
2. Read the vertex (h, k).
3. Find the signed displacement p from the vertex to the focus along the axis.
4. Substitute into the appropriate standard form.

Worked example 4

A parabola has vertex (2, −1) and focus (2, 2).

Find its equation.

The focus is directly above the vertex, so the axis is vertical.

\[ p=2-(-1)=3. \]

Use:

\[ (x-h)^2=4p(y-k). \]

Therefore:

\[ (x-2)^2=12(y+1). \]

The directrix is:

\[ y=k-p=-1-3=-4. \]

Check: The vertex lies halfway between the focus at y = 2 and the directrix at y = −4.

Method B: Given the focus and directrix

1. Draw or identify the perpendicular axis through the focus.
2. Find the point where that axis meets the directrix.
3. Take the midpoint of that point and the focus to obtain the vertex.
4. Find p and write the equation.

Worked example 5

A parabola has focus (−2, 3) and directrix x = 4.

Find its equation.

The directrix is vertical, so the axis is horizontal:

Axis: y = 3.

The point where the axis meets the directrix is (4, 3).

The vertex is the midpoint of (−2, 3) and (4, 3):

Vertex: (1, 3).

The signed displacement from the vertex to the focus is:

\[ p=-2-1=-3. \]

Therefore:

\[ (y-3)^2=-12(x-1). \]

The parabola opens left, toward the focus.

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10. Completing the square

An equation may need rearranging before its geometric properties become visible.

Worked example 6

Find the vertex, focus, and directrix of:

\[ y^2-6y-8x+17=0. \]

Step 1: Rearrange

\[ y^2-6y=8x-17. \]

Step 2: Complete the square

Add 9 to both sides:

\[ y^2-6y+9=8x-8. \]

Therefore:

\[ (y-3)^2=8(x-1). \]

Step 3: Read the parameters

\[ h=1,\qquad k=3,\qquad 4p=8. \]\[ p=2. \]

Therefore:

• Vertex: (1, 3).
• Focus: (3, 3).
• Directrix: x = −1.
• Axis of symmetry: y = 3.
• Opening direction: right.

Check: Expanding the standard form reproduces the original equation.

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11. Connecting parabolas with quadratic functions

A quadratic function:

\[ y=Ax^2+Bx+C, \qquad A\neq0, \]

has a vertical parabola as its graph.

After completing the square, it can be written as:

\[ y=A(x-h)^2+k. \]

Rearrange:

\[ (x-h)^2=\frac1A(y-k). \]

Compare with:

\[ (x-h)^2=4p(y-k). \]

Therefore:

\[ 4p=\frac1A, \qquad p=\frac1{4A}. \]

Important distinction

The coefficient A in a quadratic function is not the focal parameter p.

Worked example 7

Find the vertex, focus, and directrix of:

\[ y=2x^2-8x+5. \]

Complete the square:

\[ y=2(x^2-4x)+5. \]\[ y=2(x-2)^2-3. \]

Therefore:

\[ (x-2)^2=\frac12(y+3). \]

Thus:

\[ 4p=\frac12, \qquad p=\frac18. \]

Vertex: (2, −3).

Focus:

\[ \left(2,-3+\frac18\right) =\left(2,-\frac{23}{8}\right). \]

Directrix:

\[ y=-3-\frac18=-\frac{25}{8}. \]

Axis of symmetry: x = 2.

The positive coefficient A means the parabola opens upward.

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12. The latus rectum

The latus rectum is the chord through the focus that is perpendicular to the axis of symmetry.

Its length is:

\[ 4|p|. \]

This provides a useful connection between the equation and the width of the parabola near its focus.

Horizontal form

For:

\[ y^2=4px, \]

the focus is (p, 0).

Set x = p:

\[ y^2=4p^2. \]

Therefore:

\[ y=\pm2|p|. \]

The endpoints are:

(p, 2|p|) and (p, −2|p|).

Vertical form

For:

\[ x^2=4py, \]

the endpoints are:

(2|p|, p) and (−2|p|, p).

Worked example 8

Find the latus rectum length and endpoints for:

\[ y^2=8x. \]

Since:

\[ 4p=8, \]

we have p = 2.

Length: 8 units.

Endpoints: (2, 4), (2, −4).

For translated parabolas, shift these coordinates by the vertex coordinates.

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13. Eccentricity and the distance definition

For a parabola, the ratio:

\[ \frac{\text{distance to the focus}} {\text{distance to the directrix}} \]

equals 1 at every point on the curve.

Therefore, the eccentricity of every parabola is:

\[ e=1. \]

Compare:

• Noncircular ellipse: 0 < e < 1.
• Parabola: e = 1.
• Hyperbola: e > 1.

A parabola has no center of symmetry and no straight-line asymptotes.

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14. Sketching and coordinate limits

Sketching procedure

1. Rewrite the equation in standard form.
2. Identify the vertex.
3. Determine the axis and opening direction.
4. Plot the focus.
5. Draw the directrix as a dashed line.
6. Find a pair of symmetric points by substitution.
7. Draw a smooth curve through the vertex and points.

Coordinate limits

For a vertical parabola with vertex (h, k):

• Opens upward: x can be any real number; y ≥ k.
• Opens downward: x can be any real number; y ≤ k.

For a horizontal parabola with vertex (h, k):

• Opens right: x ≥ h; y can be any real number.
• Opens left: x ≤ h; y can be any real number.

A vertical parabola is the graph of a function y = f(x).

A complete horizontal parabola is not: most permitted x-values correspond to two y-values.

Width comparison

For parabolas with the same vertex and axis, a larger |p| produces a wider opening.

Equivalently, in:

\[ y=A(x-h)^2+k, \]

a smaller |A| produces a wider graph.

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15. Checking points and finding intersections

Worked example 9: Checking a point

Does P = (2, 4) lie on:

\[ y^2=8x? \]

Substitute:

\[ 4^2=8(2). \]

Both sides equal 16.

Therefore, P lies on the parabola.

Worked example 10: Intersecting a line

Find all intersection points of:

\[ y=2x \]

and:

\[ y^2=8x. \]

Substitute y = 2x:

\[ (2x)^2=8x. \]\[ 4x^2-8x=0. \]

Factor:

\[ 4x(x-2)=0. \]

Therefore:

\[ x=0\quad\text{or}\quad x=2. \]

Using y = 2x:

• When x = 0, y = 0.
• When x = 2, y = 4.

Intersection points: (0, 0), (2, 4).

Important: Dividing by x would lose the solution x = 0. Factor instead.

Number of intersections

Substitution can produce:

• Two distinct real solutions: two intersection points.
• One repeated real solution: one intersection point, typically tangency in the quadratic substitution case.
• No real solutions: no intersection.

A line parallel to the parabola’s axis can also meet it once without being tangent. Therefore, “one intersection” alone does not always imply tangency.

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16. Using the distance definition directly

Sometimes the focus-distance question can be solved without calculating a square root.

Worked example 11

A point P on:

\[ y^2=12x \]

has x-coordinate 5.

Find its distance to the focus.

The focus is (3, 0), and the directrix is x = −3.

By the parabola definition, the focal distance equals the perpendicular distance to the directrix:

\[ PF=|5-(-3)|=8. \]

Distance to the focus: 8 units.

The y-coordinate is unnecessary.

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17. Common mistakes and verification

Mistake 1: Reading the coefficient as p

In y² = 12x:

\[ 4p=12, \]

so p = 3, not 12.

Mistake 2: Squaring the wrong variable

• y² = 4px has a horizontal axis.
• x² = 4py has a vertical axis.

Mistake 3: Ignoring the sign of p

A negative p reverses the opening direction.

Distances remain nonnegative, so the focal length is |p|.

Mistake 4: Putting the focus and directrix on the same side

They lie on opposite sides of the vertex, at equal perpendicular distances.

Mistake 5: Confusing the axis and directrix

The axis passes through the focus and vertex.

The directrix is perpendicular to that axis.

Mistake 6: Reading translated signs incorrectly

In:

\[ (y+2)^2=8(x-3), \]

the vertex is (3, −2).

Mistake 7: Using A as p in a quadratic function

For y = A(x − h)² + k:

\[ p=\frac1{4A}. \]

Mistake 8: Calling every equation with one squared variable a parabola

For example:

\[ x^2=4 \]

represents two vertical lines.

A nondegenerate axis-aligned parabola requires the other variable to appear linearly with a nonzero coefficient.

Verification checklist

• Is the equation in standard form?
• Have you divided the coefficient by 4 to find p?
• Is the focus on the opening side?
• Is the vertex halfway between the focus and directrix?
• Is the axis perpendicular to the directrix?
• Does the vertex satisfy the equation?
• Do substituted test points satisfy the equation?

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18. Independent practice

Complete these before reading the solutions.

Practice 1

For:

\[ y^2=20x, \]

find the vertex, focus, directrix, axis, opening direction, and latus rectum length.

Practice 2

Find the focus and directrix of:

\[ x^2=-12y. \]

Practice 3

Find the equation of the parabola with vertex (0, 0) and focus (−4, 0).

Practice 4

A parabola has focus (0, 3) and directrix y = −3.

Find its equation.

Practice 5

Find the vertex, focus, directrix, and axis of:

\[ (x-2)^2=8(y+1). \]

Practice 6

Find the equation of the parabola with vertex (−1, 2) and focus (−1, −1).

Practice 7

Convert to standard form and identify the focus:

\[ x^2+4x-12y+16=0. \]

Practice 8

Find the focus and directrix of:

\[ y=\frac14(x+2)^2-3. \]

Practice 9

Find all intersection points of y = x with:

\[ y^2=4x. \]

Practice 10

A point on:

\[ x^2=8y \]

has y-coordinate 5.

Find its distance to the focus.

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19. Practice solutions

Solution 1

\[ 4p=20,\qquad p=5. \]

• Vertex: (0, 0).
• Focus: (5, 0).
• Directrix: x = −5.
• Axis: y = 0.
• Opening direction: right.
• Latus rectum length: 20.

Solution 2

\[ 4p=-12,\qquad p=-3. \]

Focus: (0, −3).

Directrix: y = 3.

Solution 3

The axis is horizontal and p = −4.

\[ y^2=-16x. \]

Solution 4

The vertex is halfway between the focus and directrix:

Vertex: (0, 0).

With p = 3:

\[ x^2=12y. \]

Solution 5

\[ h=2,\qquad k=-1,\qquad p=2. \]

• Vertex: (2, −1).
• Focus: (2, 1).
• Directrix: y = −3.
• Axis: x = 2.

Solution 6

The focus is 3 units below the vertex:

\[ p=-3. \]

Therefore:

\[ (x+1)^2=-12(y-2). \]

Solution 7

Rearrange:

\[ x^2+4x=12y-16. \]

Complete the square:

\[ (x+2)^2=12y-12. \]\[ (x+2)^2=12(y-1). \]

The vertex is (−2, 1), and p = 3.

Focus: (−2, 4).

Solution 8

Rearrange:

\[ (x+2)^2=4(y+3). \]

Thus p = 1 and the vertex is (−2, −3).

Focus: (−2, −2).

Directrix: y = −4.

Solution 9

Substitute y = x:

\[ x^2=4x. \]\[ x(x-4)=0. \]

Therefore x = 0 or x = 4.

Intersection points: (0, 0), (4, 4).

Solution 10

\[ 4p=8,\qquad p=2. \]

The directrix is y = −2.

The point’s distance to the focus equals its distance to the directrix:

\[ PF=|5-(-2)|=7. \]

Distance: 7 units.

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20. Lesson assessment — 03.6 Parabolas Quiz

Suggested time: 15 minutes
Questions: 10 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 8/10

Question 1

Find the focus of:

\[ y^2=16x. \]

A. (8, 0)
B. (0, 4)
C. (4, 0)
D. (−4, 0)

Question 2

Find the directrix of:

\[ x^2=-20y. \]

A. y = 5
B. y = −5
C. x = 5
D. y = 20

Question 3

Which equation represents a parabola opening left?

A. \(x^2=-8y\)

B. \(y^2=8x\)

C. \(x^2=8y\)

D. \(y^2=-8x\)

Question 4

Find the vertex of:

\[ (y-3)^2=12(x+2). \]

A. (2, 3)
B. (−2, 3)
C. (3, −2)
D. (−2, −3)

Question 5

A parabola has focus (2, 1) and directrix x = −2.

Which is its equation?

A. \((x-2)^2=8(y-1)\)

B. \((y-1)^2=4x\)

C. \((y-1)^2=8x\)

D. \((y+1)^2=8x\)

Question 6

Find the focus of:

\[ (x+1)^2=-8(y-2). \]

A. (−1, 0)
B. (−1, 4)
C. (1, 0)
D. (−3, 2)

Question 7

Find the focus of:

\[ y^2+4y-12x+16=0. \]

A. (1, −2)
B. (−2, 4)
C. (3, −2)
D. (4, −2)

Question 8

Find the latus rectum length of:

\[ x^2=6y. \]

A. 3/2
B. 6
C. 3
D. 12

Question 9

Find all intersection points of y = x with:

\[ y^2=8x. \]

A. (0, 0) only
B. (8, 8) only
C. (0, 0), (8, 8)
D. (0, 0), (−8, −8)

Question 10

A point P on:

\[ y^2=12x \]

has x-coordinate 6.

What is its distance to the focus?

A. 9
B. 3
C. 6
D. 12

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21. Instructor answer key and assessment solutions

Tutor LMS placement: Put the assessment in a separate Quiz item. Keep this answer key and the explanations separate from student-facing content until submission.