Chapter 06 — Coordinate Geometry: Part 7 — Conic Sections: The Hyperbola

MATHEMATICS HIPHOP ACADEMY

CHAPTER 06: COORDINATE GEOMETRY

PART 7 — CONIC SECTIONS: THE HYPERBOLA

A-Level Pure Mathematics Paper 1

Learning objectives

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

• Define a hyperbola and describe its important features.
• Recognise horizontal and vertical hyperbolas from their equations.
• Find the centre, vertices, foci and eccentricity.
• Determine the equations of asymptotes and directrices.
• Find the equation of a hyperbola from given information.
• Use completing the square to obtain standard form.
• Find tangents and normals to hyperbolas.
• Solve examination-style problems involving hyperbolas.

1. Introduction to the Hyperbola

A hyperbola is a conic section consisting of two separate branches. It is the locus of a point whose absolute difference in distances from two fixed points, called the foci, is constant.

If the two foci are F₁ and F₂ and P is any point on the hyperbola, then:

|PF₁ − PF₂| = constant

The constant difference is equal to twice the distance from the centre to a vertex, that is, 2a.

Unlike an ellipse, a hyperbola has two separate branches. Each branch extends indefinitely and approaches certain straight lines called asymptotes.

1.1 The main parts of a hyperbola

• Centre: The midpoint between the two foci and the two vertices.
• Foci: The two fixed points used in defining the hyperbola.
• Vertices: The points where the branches are closest to the centre.
• Transverse axis: The axis passing through the centre and both vertices.
• Conjugate axis: The axis through the centre perpendicular to the transverse axis.
• Asymptotes: Straight lines that the branches approach as they extend indefinitely.
• Eccentricity: A measure of the shape of the hyperbola.
• Directrices: Fixed lines associated with the focus-directrix definition.

1.2 The eccentricity of a hyperbola

The eccentricity of a hyperbola is always greater than 1.

e > 1

This distinguishes a hyperbola from an ellipse, whose eccentricity is less than 1, and a parabola, whose eccentricity is exactly 1.

2. Standard Equation of a Horizontal Hyperbola

A hyperbola whose transverse axis is parallel to the x-axis has the standard equation:

x²/a² − y²/b² = 1

Centre at the origin (0, 0)

Here, a and b are positive constants.

2.1 Important properties

• Centre: (0, 0)
• Vertices: (−a, 0) and (a, 0)
• Foci: (−c, 0) and (c, 0)
• Transverse axis length: 2a
• Conjugate axis length: 2b

The relationship between a, b and c is:

c² = a² + b²

Notice that this relationship is different from the corresponding relationship for an ellipse, where c² = a² − b².

2.2 Asymptotes

The asymptotes of a horizontal hyperbola centred at the origin are:

y = ±(b/a)x

Therefore, the two equations are:

• y = (b/a)x
• y = −(b/a)x

2.3 Eccentricity

For a horizontal hyperbola:

e = c/a

Therefore, e = √(a² + b²)/a

2.4 Directrices

The directrices of a horizontal hyperbola centred at the origin are:

x = ±a/e

3. Standard Equation of a Vertical Hyperbola

A hyperbola whose transverse axis is parallel to the y-axis has the equation:

y²/a² − x²/b² = 1

Centre at the origin (0, 0)

3.1 Important properties

• Centre: (0, 0)
• Vertices: (0, −a) and (0, a)
• Foci: (0, −c) and (0, c)
• Transverse axis length: 2a
• Conjugate axis length: 2b

As before:

c² = a² + b²

3.2 Asymptotes

The asymptotes are:

y = ±(a/b)x

3.3 Eccentricity and directrices

The eccentricity is:

e = c/a

The directrices are:

y = ±a/e

4. Hyperbolas with a Centre Other Than the Origin

A hyperbola may be translated so that its centre is at (h, k), rather than at the origin.

4.1 Horizontal hyperbola

(x − h)²/a² − (y − k)²/b² = 1

Its main properties are:

• Centre: (h, k)
• Vertices: (h ± a, k)
• Foci: (h ± c, k)
• Asymptotes: y − k = ±(b/a)(x − h)
• Directrices: x = h ± a/e

4.2 Vertical hyperbola

(y − k)²/a² − (x − h)²/b² = 1

Its main properties are:

• Centre: (h, k)
• Vertices: (h, k ± a)
• Foci: (h, k ± c)
• Asymptotes: y − k = ±(a/b)(x − h)
• Directrices: y = k ± a/e

4.3 How to identify the direction

Look at the positive squared term in the standard equation:

• If the x-term is positive, the hyperbola opens left and right.
• If the y-term is positive, the hyperbola opens upwards and downwards.

Important: The positive term determines the direction of the transverse axis.

5. Completing the Square to Find Standard Form

Some hyperbola equations are given in expanded form. Completing the square helps us rewrite them in standard form.

Consider:

x² − 4y² − 6x + 8y − 3 = 0

Step 1: Group the x-terms and y-terms.

(x² − 6x) − 4(y² − 2y) = 3

Step 2: Complete the square.

x² − 6x = (x − 3)² − 9

y² − 2y = (y − 1)² − 1

Step 3: Substitute.

(x − 3)² − 9 − 4[(y − 1)² − 1] = 3

Step 4: Simplify.

(x − 3)² − 9 − 4(y − 1)² + 4 = 3

(x − 3)² − 4(y − 1)² = 8

Step 5: Divide through by 8.

(x − 3)²/8 − (y − 1)²/2 = 1

Comparing with standard form:

• Centre: (3, 1)
• a² = 8, so a = 2√2
• b² = 2, so b = √2
• c² = 8 + 2 = 10, so c = √10
• Horizontal transverse axis

6. The Latus Rectum

The latus rectum is a chord passing through a focus and perpendicular to the transverse axis.

For either orientation of a standard hyperbola, the length of the latus rectum is:

Length = 2b²/a

For a horizontal hyperbola, the latus rectum through the right focus x = c has endpoints:

(c, b²/a) and (c, −b²/a).

The corresponding chord through the left focus has x-coordinate −c.

7. Parametric Equations

A convenient parametrisation of the horizontal hyperbola is:

x = a sec θ

y = b tan θ

To verify this, substitute into the standard equation:

x²/a² − y²/b² = sec²θ − tan²θ

Using the identity sec²θ − tan²θ = 1, we obtain:

x²/a² − y²/b² = 1.

For the vertical hyperbola, one possible parametrisation is:

x = b tan θ

y = a sec θ

8. Tangents to a Hyperbola

A tangent is a straight line that touches a curve at a point and has the same gradient as the curve at that point.

8.1 Tangent to a horizontal hyperbola

For the hyperbola:

x²/a² − y²/b² = 1

The tangent at a point (x₁, y₁) on the curve is:

xx₁/a² − yy₁/b² = 1

The point must lie on the hyperbola.

8.2 Tangent to a vertical hyperbola

For:

y²/a² − x²/b² = 1

The tangent at (x₁, y₁) is:

yy₁/a² − xx₁/b² = 1

8.3 Finding a tangent using differentiation

Consider:

x²/a² − y²/b² = 1

Differentiating implicitly with respect to x:

2x/a² − (2y/b²)(dy/dx) = 0

Rearranging:

dy/dx = b²x/(a²y)

Therefore, at (x₁, y₁), the gradient of the tangent is:

m = b²x₁/(a²y₁).

Using the point-gradient equation:

y − y₁ = m(x − x₁).

9. Normals to a Hyperbola

A normal is a straight line perpendicular to the tangent at the point of contact.

If the gradient of the tangent is m, then the gradient of the normal is −1/m, provided m is non-zero.

For the horizontal hyperbola:

m_tangent = b²x₁/(a²y₁)

Therefore:

m_normal = −a²y₁/(b²x₁)

The normal at (x₁, y₁) is:

y − y₁ = −a²y₁/(b²x₁)(x − x₁).

This form assumes x₁ is non-zero. If the tangent is horizontal or vertical, handle the perpendicular line directly.

10. The Rectangular Hyperbola

A rectangular hyperbola is a special type of hyperbola whose asymptotes are perpendicular to each other.

For a horizontal hyperbola, the asymptotes have gradients ±b/a. They are perpendicular when:

(b/a)(−b/a) = −1

This gives b²/a² = 1, so a = b.

Its standard equation is therefore:

x²/a² − y²/a² = 1

Multiplying through by a²:

x² − y² = a²

Another familiar rectangular hyperbola is xy = k, where k is a non-zero constant. Its coordinate axes are its asymptotes.

11. Worked Examples

Example 1: Finding the features of a hyperbola

Find the centre, vertices, foci, eccentricity and asymptotes of:

x²/25 − y²/9 = 1

Solution

Comparing with x²/a² − y²/b² = 1:

a² = 25, so a = 5.

b² = 9, so b = 3.

Find c:

c² = a² + b² = 25 + 9 = 34.

Therefore c = √34.

Centre: (0, 0)

Vertices: (−5, 0), (5, 0)

Foci: (−√34, 0), (√34, 0)

Eccentricity:

e = c/a = √34/5.

Asymptotes:

y = ±(3/5)x.

Example 2: Finding an equation from the vertices and foci

A hyperbola has vertices at (−3, 0) and (3, 0), and foci at (−5, 0) and (5, 0). Find its equation.

Solution

The centre is (0, 0), and the transverse axis is horizontal.

a = 3 and c = 5.

Using c² = a² + b²:

25 = 9 + b²

b² = 16.

Substitute into the standard equation:

x²/9 − y²/16 = 1

Example 3: Finding the features of a vertical hyperbola

Given:

y²/16 − x²/9 = 1

Find its centre, vertices, foci, eccentricity and asymptotes.

Solution

a² = 16, so a = 4.

b² = 9, so b = 3.

c² = 16 + 9 = 25, so c = 5.

Centre: (0, 0)

Vertices: (0, −4), (0, 4)

Foci: (0, −5), (0, 5)

Eccentricity: e = 5/4.

Asymptotes:

y = ±(4/3)x.

Example 4: Finding the equation of a translated hyperbola

A horizontal hyperbola has centre (2, −1), a = 4 and b = 3. Find its equation, foci and asymptotes.

Solution

The standard equation is:

(x − h)²/a² − (y − k)²/b² = 1.

Substituting h = 2, k = −1, a² = 16 and b² = 9:

(x − 2)²/16 − (y + 1)²/9 = 1

Find c:

c² = 16 + 9 = 25, so c = 5.

Foci: (2 − 5, −1) and (2 + 5, −1).

Therefore, the foci are (−3, −1) and (7, −1).

Asymptotes:

y + 1 = ±(3/4)(x − 2).

Example 5: Finding a tangent

Find the equation of the tangent to:

x²/16 − y²/9 = 1

at the point (5, 9/4).

Step 1: Check that the point lies on the hyperbola.

25/16 − (81/16)/9 = 25/16 − 9/16 = 1.

The point is on the hyperbola.

Step 2: Use the tangent formula.

xx₁/a² − yy₁/b² = 1.

Substitute x₁ = 5, y₁ = 9/4, a² = 16 and b² = 9:

5x/16 − (9y/4)/9 = 1

5x/16 − y/4 = 1.

Multiply by 16:

5x − 4y = 16

Example 6: Finding the latus rectum

Find the length of the latus rectum of:

x²/36 − y²/16 = 1.

Solution

a² = 36, so a = 6.

b² = 16.

Length of latus rectum = 2b²/a.

= 2(16)/6

= 16/3 units

12. Applications of Hyperbolas

Hyperbolas are useful in mathematics, science, engineering and technology.

• Navigation: Differences in distances from fixed reference points can be used in certain positioning methods.
• Astronomy: Hyperbolic paths can describe the trajectories of objects that pass a central body and do not remain in a closed orbit.
• Optics: Hyperbolic mirrors and reflecting surfaces are used in some optical systems.
• Engineering: Hyperbolic curves appear in the design and analysis of some structures and systems.
• Mathematical modelling: Hyperbolas occur in relationships involving inverse variation, such as xy = k.

13. Examination Practice

Attempt the following questions before opening the solutions. Show all your working clearly.

Exercise A: Basic properties

Question 1. Given the hyperbola x²/49 − y²/25 = 1, find:

1. The centre.
2. The vertices.
3. The foci.
4. The eccentricity.
5. The equations of the asymptotes.

Question 2. Find the centre, vertices, foci and asymptotes of:

y²/36 − x²/28 = 1.

Question 3. A hyperbola has a = 5 and b = 12. Find c and its eccentricity.

Exercise B: Finding equations

Question 1. Find the equation of a horizontal hyperbola centred at the origin whose vertices are (±4, 0) and whose foci are (±6, 0).

Question 2. A vertical hyperbola has centre (1, −2), a = 3 and b = 4. Find its equation and foci.

Question 3. Find the standard form of:

x² − 4y² − 6x + 8y − 3 = 0.

Hence find its centre and values of a², b² and c².

Exercise C: Asymptotes and latus rectum

Question 1. Find the equations of the asymptotes of:

x²/64 − y²/36 = 1.

Question 2. Find the length of the latus rectum of:

y²/25 − x²/144 = 1.

Question 3. A hyperbola has equation:

(x − 3)²/16 − (y + 2)²/9 = 1.

Find its centre, vertices, foci and asymptotes.

Exercise D: Tangents and differentiation

Question 1. Find the gradient of the tangent to:

x²/25 − y²/9 = 1

at the point (5, 0).

Question 2. Find the equation of the tangent to:

x²/9 − y²/4 = 1

at the point (3√2, 2).

Question 3. By implicit differentiation, find dy/dx for:

x²/16 − y²/9 = 1.

Exercise E: Examination-style problems

Question 1. The foci of a horizontal hyperbola are (−5, 0) and (5, 0). Its eccentricity is 5/3. Find:

1. The value of a.
2. The value of b².
3. The equation of the hyperbola.
4. The equations of its asymptotes.

Question 2. A hyperbola has equation:

9x² − 16y² − 54x − 64y − 127 = 0.

Express the equation in standard form and determine its centre, orientation, vertices and foci.

Question 3. A rectangular hyperbola has equation xy = 12. Find its intercept-free asymptotic behaviour and explain why its coordinate axes are asymptotes.

14. Answers and Worked Solutions

Use these solutions to mark your work and understand any mistakes.

Exercise A — Solutions

Question 1.

a = 7, b = 5 and c² = 49 + 25 = 74, so c = √74.

• Centre: (0, 0)
• Vertices: (±7, 0)
• Foci: (±√74, 0)
• e = √74/7
• Asymptotes: y = ±(5/7)x

Question 2.

a² = 36, b² = 28, so c² = 64 and c = 8.

• Centre: (0, 0)
• Vertices: (0, ±6)
• Foci: (0, ±8)
• Asymptotes: y = ±(6/√28)x = ±(3/√7)x

Question 3.

c² = 25 + 144 = 169, so c = 13.

e = 13/5.

Exercise B — Solutions

Question 1.

a = 4, c = 6.

b² = c² − a² = 36 − 16 = 20.

Equation: x²/16 − y²/20 = 1.

Question 2.

a² = 9 and b² = 16.

Equation: (y + 2)²/9 − (x − 1)²/16 = 1.

c² = 9 + 16 = 25, so c = 5.

Foci: (1, −2 ± 5), which are (1, 3) and (1, −7).

Question 3.

Group the terms:

(x² − 6x) − 4(y² − 2y) = 3.

Complete the square:

(x − 3)² − 9 − 4[(y − 1)² − 1] = 3.

(x − 3)² − 4(y − 1)² = 8.

Divide by 8:

(x − 3)²/8 − (y − 1)²/2 = 1.

• Centre: (3, 1)
• a² = 8
• b² = 2
• c² = 10

Exercise C — Solutions

Question 1.

a = 8, b = 6.

Asymptotes: y = ±(3/4)x.

Question 2.

a = 5 and b² = 144.

Length = 2b²/a = 288/5 units.

Question 3.

Centre: (3, −2).

a = 4, b = 3 and c = 5.

Vertices: (−1, −2) and (7, −2).

Foci: (−2, −2) and (8, −2).

Asymptotes: y + 2 = ±(3/4)(x − 3).

Exercise D — Solutions

Question 1.

For x²/25 − y²/9 = 1:

dy/dx = 9x/(25y).

At (5, 0), the denominator is zero. The tangent is vertical, namely x = 5. Its gradient is undefined.

Question 2.

First check the point:

(3√2)²/9 − 2²/4 = 18/9 − 1 = 1.

It lies on the hyperbola.

Using the tangent formula:

(3√2)x/9 − 2y/4 = 1.

Simplifying:

2√2 x − 3y = 6.

Question 3.

Differentiate:

2x/16 − (2y/9)(dy/dx) = 0.

Therefore:

dy/dx = 9x/(16y).

Exercise E — Solutions

Question 1.

c = 5 and e = 5/3.

Since e = c/a:

5/3 = 5/a, so a = 3.

b² = c² − a² = 25 − 9 = 16.

Equation: x²/9 − y²/16 = 1.

Asymptotes: y = ±(4/3)x.

Question 2.

Group and complete the square:

9(x² − 6x) − 16(y² + 4y) = 127.

9[(x − 3)² − 9] − 16[(y + 2)² − 4] = 127.

9(x − 3)² − 81 − 16(y + 2)² + 64 = 127.

9(x − 3)² − 16(y + 2)² = 144.

Divide by 144:

(x − 3)²/16 − (y + 2)²/9 = 1.

Centre: (3, −2).

a = 4, b = 3 and c = 5.

Orientation: horizontal.

Vertices: (−1, −2) and (7, −2).

Foci: (−2, −2) and (8, −2).

Question 3.

For xy = 12, write y = 12/x. As x approaches positive or negative infinity, y approaches zero. As x approaches zero, the magnitude of y becomes arbitrarily large.

The coordinate axes x = 0 and y = 0 are asymptotes. The curve approaches these axes but does not meet either one because xy = 12 cannot hold when x = 0 or y = 0.

15. Summary of Important Formulae

Feature	Horizontal	Vertical
Standard form	x²/a² − y²/b² = 1	y²/a² − x²/b² = 1
Centre	(0, 0)	(0, 0)
Vertices	(±a, 0)	(0, ±a)
Foci	(±c, 0)	(0, ±c)
Focal relation	c² = a² + b²
Eccentricity	e = c/a, where e > 1
Asymptotes	y = ±(b/a)x	y = ±(a/b)x
Directrices	x = ±a/e	y = ±a/e
Latus rectum	2b²/a

Final examination reminders

1. Always identify the positive squared term before deciding the orientation.
2. Remember that c² = a² + b² for a hyperbola.
3. Do not confuse the hyperbola's asymptotes with its axes.
4. When finding a tangent at a given point, first verify that the point lies on the curve.
5. When completing the square, remember to account for any coefficient outside the bracket.
6. Keep exact values in surd form unless a decimal approximation is requested.

MATHEMATICS HIPHOP

Making Mathematics Easier, One Concept at a Time.

End of Chapter 06 — Part 7