Chapter 06 — Coordinate Geometry Part 5: Conic Sections — Introduction to Parabolas

CHAPTER 06 — COORDINATE GEOMETRY

PART 5: CONIC SECTIONS — INTRODUCTION TO PARABOLAS

Learning objectives

By the end of this lesson, you should be able to:

• Explain the meaning of a conic section and define a parabola.
• Identify the vertex, focus, directrix and axis of symmetry.
• Derive and use the standard equations of a parabola.
• Sketch parabolas and identify their important features.
• Determine the equation of a parabola from given information.
• Use parametric equations to represent points on a parabola.
• Find equations of tangents, normals and chords of a parabola.
• Determine the length of the latus rectum.

1. Introduction to Conic Sections

A conic section is a curve obtained when a plane cuts through a double cone at different angles. The main conic sections are:

1. Circle: formed when the cutting plane is perpendicular to the axis of the cone.
2. Ellipse: a closed oval-shaped curve.
3. Parabola: an open curve formed when the plane is parallel to a sloping side of the cone.
4. Hyperbola: a curve consisting of two separate branches.

Important idea

In this lesson, we concentrate on the parabola. Ellipses and hyperbolas can be studied in subsequent lessons.

2. Meaning and Definition of a Parabola

A parabola is the locus of a point which moves so that its distance from a fixed point is always equal to its perpendicular distance from a fixed straight line. The fixed point is called the focus, while the fixed straight line is called the directrix. If P is any point on a parabola, F is the focus and the perpendicular distance from P to the directrix is PM, then:

PF = PM

This property is the fundamental definition of a parabola.

Important parts of a parabola

• Vertex: the turning point of the parabola.
• Focus: the fixed point used in defining the parabola.
• Directrix: the fixed straight line used in defining the parabola.
• Axis of symmetry: the straight line passing through the vertex and focus, perpendicular to the directrix.
• Latus rectum: a chord passing through the focus and perpendicular to the axis of symmetry.

Illustration: A parabola opening to the right

Schematic illustration only. The vertex is V, the focus is F, and the vertical dashed line represents the directrix.

3. Derivation of the Standard Equation

Consider a parabola with:

• Vertex at the origin O(0,0).
• Focus F(a,0).
• Directrix x = −a.
• Axis of symmetry along the x-axis.

Let P(x,y) be any point on the parabola. By the definition of a parabola: PF = perpendicular distance from P to the directrix. Using the distance formula:

PF = √[(x − a)² + y²]

The perpendicular distance from P(x,y) to x = −a is:

PM = x + a

Therefore: √[(x − a)² + y²] = x + a Squaring both sides: (x − a)² + y² = (x + a)² Expanding: x² − 2ax + a² + y² = x² + 2ax + a² Simplifying: y² = 4ax Hence, the standard equation is:

y² = 4ax

For a parabola opening to the right, a is positive.

4. Standard Forms of a Parabola

The orientation of a parabola depends on the position of its axis of symmetry.

Equation	Direction	Focus	Directrix
y² = 4ax	Right	(a,0)	x = −a
y² = −4ax	Left	(−a,0)	x = a
x² = 4ay	Upwards	(0,a)	y = −a
x² = −4ay	Downwards	(0,−a)	y = a

In all four standard forms, the vertex is at the origin.

Understanding the parameter a

The positive quantity a represents the distance between the vertex and the focus. For example, in y² = 20x: 4a = 20 a = 5. The focus is therefore (5,0), and the directrix is x = −5.

5. The Latus Rectum

The latus rectum is a chord passing through the focus and perpendicular to the axis of symmetry. For the parabola: y² = 4ax the focus is (a,0). The latus rectum is therefore the vertical line: x = a. Substituting x = a into the parabola: y² = 4a(a) y² = 4a² y = ±2a. The endpoints of the latus rectum are:

(a, 2a) and (a, −2a)

Its length is:

Length of latus rectum = 4a

The same length, 4a, applies to all four standard orientations when a is taken as a positive distance.

6. Translated Parabolas

Not every parabola has its vertex at the origin. If the vertex is (h,k), the standard equations become:

(y − k)² = 4a(x − h)

This parabola opens to the right when a is positive. Its:

• Vertex is (h,k).
• Focus is (h + a,k).
• Directrix is x = h − a.
• Axis of symmetry is y = k.

For a vertical parabola:

(x − h)² = 4a(y − k)

Its:

• Vertex is (h,k).
• Focus is (h,k + a).
• Directrix is y = k − a.
• Axis of symmetry is x = h.

Worked Example 1

Find the vertex, focus, directrix and axis of symmetry of: (y − 3)² = 12(x − 2). Solution Compare with: (y − k)² = 4a(x − h). Therefore: h = 2, k = 3 and 4a = 12. a = 3. Vertex = (2,3). Focus = (h + a,k) = (5,3). Directrix: x = h − a x = 2 − 3 = −1. Axis of symmetry: y = 3. Answer:

• Vertex: (2,3)
• Focus: (5,3)
• Directrix: x = −1
• Axis: y = 3

7. Finding the Equation from Given Information

A parabola's equation can be determined from its vertex and focus, or from its vertex and directrix.

Worked Example 2

Find the equation of a parabola whose vertex is at the origin and whose focus is (4,0). Solution The focus is (a,0). Therefore a = 4. Using: y² = 4ax y² = 4(4)x

y² = 16x

Worked Example 3

Find the equation of a parabola with vertex (1,−2) and focus (1,2). Solution The vertex and focus have the same x-coordinate. Therefore the axis is vertical. The vertex is (h,k) = (1,−2). The focus is 4 units above the vertex: a = 2(−2 to 2 gives a distance of 4) = 4. Use: (x − h)² = 4a(y − k). Substituting: (x − 1)² = 4(4)(y + 2).

(x − 1)² = 16(y + 2)

8. Parametric Equations of a Parabola

For the parabola: y² = 4ax, a point on the curve can be represented using a parameter t. Let: y = 2at. Substitute into y² = 4ax: (2at)² = 4ax 4a²t² = 4ax x = at². Therefore the parametric coordinates are:

x = at²,   y = 2at

The point is written as:

P(at², 2at)

The parameter t can take any real value.

Worked Example 4

Find the point on the parabola y² = 12x corresponding to t = 2. Solution Compare: y² = 4ax 4a = 12, so a = 3. Using: x = at² and y = 2at, x = 3(2²) = 12. y = 2(3)(2) = 12. Answer: The point is (12,12).

9. Tangents to a Parabola

A tangent is a straight line which touches a curve at a particular point. For: y² = 4ax, the equation of the tangent at a point (x₁,y₁) on the parabola is:

yy₁ = 2a(x + x₁)

If the point is given parametrically as (at²,2at), the tangent becomes:

ty = x + at²

Worked Example 5

Find the equation of the tangent to y² = 16x at the point (4,8). Solution Compare y² = 4ax with y² = 16x. 4a = 16, so a = 4. The point is (x₁,y₁) = (4,8). Using: yy₁ = 2a(x + x₁), 8y = 2(4)(x + 4). 8y = 8x + 32. Dividing by 8:

y = x + 4

10. Normals to a Parabola

A normal is a straight line perpendicular to the tangent at the point of contact. For y² = 4ax, the gradient of the tangent at the parametric point (at²,2at) is 1/t, provided t is not zero. Therefore, the gradient of the normal is −t. Using the point-gradient equation: y − 2at = −t(x − at²). Simplifying:

y = −tx + 2at + at³

This is the equation of the normal at the parametric point (at²,2at).

Worked Example 6

Find the equation of the normal to y² = 8x at the point corresponding to t = 1. Solution 4a = 8, so a = 2. The point is: x = at² = 2(1)² = 2. y = 2at = 2(2)(1) = 4. The normal is: y = −tx + 2at + at³. Substitute a = 2 and t = 1: y = −x + 4 + 2.

x + y − 6 = 0

11. Chords of a Parabola

A chord is a straight line segment joining two points on a curve. For the parabola y² = 4ax, let two points have parameters t₁ and t₂. Their coordinates are: P(at₁²,2at₁) and Q(at₂²,2at₂). The equation of the chord joining these points is:

(t₁ + t₂)y = 2x + 2at₁t₂

This formula is useful when the two points are given in parametric form.

Worked Example 7

Find the equation of the chord joining the points on y² = 12x corresponding to t = 1 and t = 3. Solution 4a = 12, hence a = 3. t₁ = 1 and t₂ = 3. Using: (t₁ + t₂)y = 2x + 2at₁t₂, (1 + 3)y = 2x + 2(3)(1)(3). 4y = 2x + 18. Dividing by 2:

x − 2y + 9 = 0

12. Applications of Parabolas

Parabolic shapes and properties are useful in many practical situations.

• Satellite dishes: a parabolic reflector helps direct signals towards a receiver near the focus.
• Reflectors: parabolic reflectors can direct light or other waves into a concentrated beam.
• Bridges and architecture: parabolic curves may be used in the design and analysis of structures.
• Projectile motion: under ideal conditions, the path of a projectile can be represented by a parabola.

Remember

A mathematical model is useful only when its assumptions are appropriate. For example, the ideal projectile path is parabolic when air resistance is neglected and gravitational acceleration is treated as constant.

13. Common Errors to Avoid

1. Confusing the focus with the vertex. They are different points.
2. Using the wrong sign in the directrix equation.
3. Forgetting that 4a, not a, is the coefficient in the standard equation.
4. Confusing a horizontal axis with a vertical axis.
5. Using the parametric equations of a parabola without first identifying a.
6. Giving the equation of a tangent or normal without checking that the specified point lies on the curve.
7. Forgetting that the latus rectum has length 4a.

14. Examination Practice

Exercise A — Basic Understanding

1. Define a parabola in terms of a fixed point and a fixed straight line.
2. State the meaning of the focus, directrix, vertex and axis of symmetry.
3. Find the focus and directrix of y² = 20x.
4. Find the vertex and focus of x² = −24y.
5. Determine the length of the latus rectum of y² = 28x.
6. State the direction in which each of the following parabolas opens: (a) y² = 12x; (b) y² = −8x; (c) x² = 16y; (d) x² = −20y.

Show answers and solutions — Exercise A

1. A parabola is the locus of a point equidistant from a fixed point and a fixed straight line.
2. Focus: fixed point; directrix: fixed line; vertex: turning point; axis: line of symmetry through the vertex and focus.
3. 4a = 20, a = 5. Focus (5,0), directrix x = −5.
4. 4a = 24, a = 6. Focus (0,−6), directrix y = 6. Vertex (0,0).
5. 4a = 28. Length = 28 units.
6. (a) Right; (b) Left; (c) Upwards; (d) Downwards.

Exercise B — Equations and Properties

1. Find the equation of a parabola with vertex at the origin and focus (6,0).
2. Find the equation of a parabola with vertex (2,1) and focus (5,1).
3. Find the vertex, focus, directrix and axis of symmetry of (y + 2)² = 20(x − 3).
4. Find the equation of a parabola whose vertex is (−1,2) and directrix is y = −3.
5. Find the endpoints and length of the latus rectum of y² = 24x.

Show answers and solutions — Exercise B

1. a = 6. Equation: y² = 24x.
2. a = 3, horizontal axis. Equation: (y − 1)² = 12(x − 2).
3. 4a = 20, a = 5. Vertex (3,−2); focus (8,−2); directrix x = −2; axis y = −2.
4. The vertex is (−1,2), and the directrix is 5 units below it. Thus a = 5. Equation: (x + 1)² = 20(y − 2).
5. a = 6. Endpoints (6,12) and (6,−12). Length = 24 units.

Exercise C — Parametric Equations, Tangents and Normals

1. For y² = 20x, find the point corresponding to t = 2.
2. Find the point on y² = 8x corresponding to t = −2.
3. Find the tangent to y² = 12x at (3,6).
4. Find the tangent to y² = 16x at the point corresponding to t = 2.
5. Find the normal to y² = 12x at the point corresponding to t = 1.
6. Find the chord joining the points corresponding to t = 2 and t = −1 on y² = 8x.

Show answers and solutions — Exercise C

1. 4a = 20, a = 5. x = 5(4) = 20; y = 2(5)(2) = 20. Point (20,20).
2. a = 2. x = 2(4) = 8; y = 2(2)(−2) = −8. Point (8,−8).
3. 4a = 12, a = 3. Using yy₁ = 2a(x + x₁): 6y = 6(x + 3). Therefore y = x + 3.
4. a = 4. Tangent ty = x + at² gives 2y = x + 16.
5. a = 3. Normal y = −tx + 2at + at³ gives y = −x + 7, or x + y − 7 = 0.
6. a = 2. (2 − 1)y = 2x + 2(2)(2)(−1). Therefore y = 2x − 8.

Exercise D — Further Examination Questions

1. A parabola has equation y² = 4ax and passes through (9,6). Find a, its focus and its directrix.
2. A parabola has vertex (2,−1) and focus (2,4). Find its equation, directrix and latus rectum length.
3. Find the equation of the tangent to y² = 20x at the point corresponding to t = 1.
4. Find the equation of the normal to y² = 16x at the point corresponding to t = −1.
5. The chord joining two points on y² = 4ax has parameters 1 and 3. Show that its equation is 4y = 2x + 6a.

Show answers and solutions — Exercise D

1. 36 = 4a(9), so a = 1. Focus (1,0); directrix x = −1.
2. The distance from vertex to focus is 5, so a = 5. Equation: (x − 2)² = 20(y + 1). Directrix y = −6. Latus rectum length = 20.
3. a = 5, t = 1. Tangent: y = x + 5.
4. a = 4, t = −1. Normal: y = x − 10.
5. Using (t₁ + t₂)y = 2x + 2at₁t₂: (1 + 3)y = 2x + 2a(1)(3). Hence 4y = 2x + 6a.

15. Summary of Important Formulae

Property	Formula
Right-opening parabola	y² = 4ax
Left-opening parabola	y² = −4ax
Upward-opening parabola	x² = 4ay
Downward-opening parabola	x² = −4ay
Focus for y² = 4ax	(a,0)
Directrix for y² = 4ax	x = −a
Latus rectum length	4a
Parametric point	(at²,2at)
Tangent at parameter t	ty = x + at²
Normal at parameter t	y = −tx + 2at + at³
Chord joining parameters t₁,t₂	(t₁+t₂)y = 2x + 2at₁t₂

Final Reminder

Always identify the orientation of a parabola before applying a formula. Check the value of a, locate the vertex and focus, and verify that any stated point lies on the curve before finding its tangent or normal.

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