Quantity	Meaning
a	Distance from the center to either vertex
b	Half the conjugate axis length; used to construct the asymptotes
c	Distance from the center to either focus
2a	Distance between the vertices; constant absolute focal-distance difference
2b	Conjugate axis length
2c	Distance between the two foci

Important: Unlike the ellipse convention, a is not necessarily larger than b.

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3. The relationship between a, b, and c

For a hyperbola:

\[ c^2=a^2+b^2. \]

Therefore:

\[ c=\sqrt{a^2+b^2}. \]

If a and c are known:

\[ b^2=c^2-a^2. \]

Compare carefully

For an ellipse:

\[ c^2=a^2-b^2. \]

For a hyperbola:

\[ c^2=a^2+b^2. \]

This difference determines whether the foci lie inside or beyond the vertices.

For a hyperbola, c > a, so the foci are farther from the center than the vertices.

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4. Standard equations centered at the origin

4.1 Horizontal transverse axis

\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \qquad a>0,\quad b>0. \]

The positive squared term contains x, so the branches open left and right.

Its properties are:

• Center: (0, 0).
• Vertices: (−a, 0), (a, 0).
• Foci: (−c, 0), (c, 0).
• Conjugate-axis endpoints: (0, −b), (0, b).

4.2 Vertical transverse axis

\[ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1, \qquad a>0,\quad b>0. \]

The positive squared term contains y, so the branches open upward and downward.

Its properties are:

• Center: (0, 0).
• Vertices: (0, −a), (0, a).
• Foci: (0, −c), (0, c).
• Conjugate-axis endpoints: (−b, 0), (b, 0).

The most useful identification rule

After making the right-hand side equal to positive 1, look at the positive squared term.

• Positive x² term: horizontal opening.
• Positive y² term: vertical opening.

The larger denominator does not determine the opening direction.

Standard-form comparison

Property	Horizontal opening	Vertical opening
Equation	x²/a² − y²/b² = 1	y²/a² − x²/b² = 1
Positive squared term	x² term	y² term
Vertices	(−a, 0), (a, 0)	(0, −a), (0, a)
Foci	(−c, 0), (c, 0)	(0, −c), (0, c)
Asymptotes	y = ±(b/a)x	y = ±(a/b)x

5. Understanding asymptotes

Horizontal hyperbola

For:

\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \]

the asymptotes are:

\[ y=\pm\frac ba x. \]

Vertical hyperbola

For:

\[ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1, \]

the asymptotes are:

\[ y=\pm\frac ab x. \]

A reliable way to obtain them

Starting from the normalized standard equation, replace the right-hand side 1 with 0.

For example:

\[ \frac{x^2}{9}-\frac{y^2}{16}=0. \]

Then:

\[ \frac{y^2}{16}=\frac{x^2}{9}. \]

Taking both signs:

\[ y=\pm\frac43x. \]

This method reduces the risk of reversing the ratio.

The construction rectangle

For a horizontal hyperbola centered at the origin:

• Move a units left and right.
• Move b units up and down.
• Form a rectangle with corners (±a, ±b).
• Extend its diagonals to obtain the asymptotes.

For a vertical hyperbola:

• Move b units left and right.
• Move a units up and down.

The rectangle is a drawing aid, not part of the hyperbola.

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6. Eccentricity

The eccentricity of a hyperbola is:

\[ e=\frac ca. \]

Since c > a:

\[ e>1. \]

Using the relationship between a, b, and c:

\[ e=\sqrt{1+\frac{b^2}{a^2}}. \]

What eccentricity tells us

Eccentricity depends on the ratio b/a, not the overall size.

Two hyperbolas with the same ratio b/a have the same eccentricity, even if their axis lengths differ.

Rectangular hyperbola

If:

\[ a=b, \]

then:

\[ e=\sqrt2. \]

The asymptotes have slopes 1 and −1, so they are perpendicular.

Such a hyperbola is called a rectangular hyperbola.

Important: Equal denominators in a difference-of-squares equation do not give a circle.

For example:

\[ x^2-y^2=9 \]

is a hyperbola.

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7. Sketching and coordinate limits

Sketching procedure

1. Normalize the equation.
2. Identify the center.
3. Identify the opening direction from the positive term.
4. Find a and b.
5. Plot the vertices.
6. Draw the construction rectangle.
7. Draw the two asymptotes through the center.
8. Sketch two smooth branches through the vertices, approaching the asymptotes.
9. Plot the foci if required.

Horizontal hyperbola centered at the origin

From:

\[ \frac{x^2}{a^2}=1+\frac{y^2}{b^2}, \]

we obtain:

\[ x^2\geq a^2. \]

Therefore:

Possible x-values: (−∞, −a] ∪ [a, ∞).

Possible y-values: all real numbers.

Vertical hyperbola centered at the origin

Similarly:

Possible x-values: all real numbers.

Possible y-values: (−∞, −a] ∪ [a, ∞).

The complete hyperbola is generally not the graph of one function y = f(x), because some x-values correspond to two y-values.

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8. Worked example 1: Reading a horizontal hyperbola

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

\[ \frac{x^2}{9}-\frac{y^2}{16}=1. \]

Step 1: Identify the direction and parameters

The positive term contains x², so the branches open left and right.

\[ a^2=9,\qquad b^2=16. \]\[ a=3,\qquad b=4. \]

Notice that b > a. This is allowed.

Step 2: Find c

\[ c^2=9+16=25. \]\[ c=5. \]

Step 3: State the points and lengths

• Center: (0, 0).
• Vertices: (−3, 0), (3, 0).
• Foci: (−5, 0), (5, 0).
• Transverse axis length: 6.
• Conjugate axis length: 8.

Step 4: Find the asymptotes

\[ y=\pm\frac43x. \]

Step 5: Find eccentricity

\[ e=\frac53. \]

Check: The eccentricity is greater than 1, and the foci lie beyond the vertices.

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9. Worked example 2: Normalize before identifying direction

Find the vertices, foci, and asymptotes of:

\[ 4x^2-9y^2=-36. \]

Step 1: Make the right-hand side positive

Multiply by −1:

\[ 9y^2-4x^2=36. \]

Divide by 36:

\[ \frac{y^2}{4}-\frac{x^2}{9}=1. \]

Step 2: Read the normalized equation

The positive term contains y², so the hyperbola opens upward and downward.

\[ a=2,\qquad b=3. \]\[ c=\sqrt{4+9}=\sqrt{13}. \]

Step 3: State the properties

Vertices: (0, −2), (0, 2).

Foci: (0, −√13), (0, √13).

Asymptotes:

\[ y=\pm\frac23x. \]

Key lesson: Reading the signs before normalizing would give the wrong opening direction.

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10. Worked example 3: Constructing an equation from vertices and foci

A hyperbola has vertices (−4, 0), (4, 0) and foci (−5, 0), (5, 0).

Find its equation.

Step 1: Identify the center and direction

The center is (0, 0).

The transverse axis is horizontal.

Step 2: Identify a and c

\[ a=4,\qquad c=5. \]

Step 3: Find b²

\[ b^2=c^2-a^2=25-16=9. \]

Step 4: Write the equation

\[ \frac{x^2}{16}-\frac{y^2}{9}=1. \]

Verification

At the vertex (4, 0):

\[ \frac{16}{16}-0=1. \]

The focus distance satisfies c > a, as required.

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11. Worked example 4: Constructing an equation from asymptotes

A hyperbola is centered at the origin, has vertices (−3, 0), (3, 0), and asymptotes:

\[ y=\pm2x. \]

Find its equation.

Step 1: Identify a and the direction

The vertices lie on the x-axis:

\[ a=3. \]

Use:

\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1. \]

Step 2: Use the asymptote slope

For a horizontal hyperbola:

\[ \frac ba=2. \]

Therefore:

\[ b=2a=6. \]

Step 3: Write the equation

\[ \frac{x^2}{9}-\frac{y^2}{36}=1. \]

Check: The asymptote slopes are ±6/3 = ±2.

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12. Worked example 5: Using eccentricity

A hyperbola is centered at the origin and opens upward and downward. Its transverse axis length is 6 and its eccentricity is 5/3.

Find its equation.

Step 1: Find a

\[ 2a=6, \]

so:

\[ a=3. \]

Step 2: Find c

\[ c=ae=3\left(\frac53\right)=5. \]

Step 3: Find b²

\[ b^2=c^2-a^2=25-9=16. \]

Step 4: Use the vertical standard form

\[ \frac{y^2}{9}-\frac{x^2}{16}=1. \]

Check: The positive term contains y², as required.

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13. Worked example 6: Focal-distance problems

For the hyperbola:

\[ \frac{x^2}{16}-\frac{y^2}{9}=1, \]

the greater of a point’s two focal distances is 13.

Find the smaller distance.

Step 1: Identify the constant difference

\[ a=4. \]

Therefore:

\[ |PF_1-PF_2|=2a=8. \]

Step 2: Use the stated order

Let the smaller distance be d.

\[ 13-d=8. \]

Smaller distance: 5 units.

Why wording matters

If the question only said “one focal distance is 13,” we could not automatically subtract 8.

The other distance could be:

• 5, if 13 is the greater distance.
• 21, if 13 is the smaller distance.

For this hyperbola, both possibilities occur.

Key lesson: An absolute difference does not specify which distance is larger.

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14. Translated hyperbolas

For center (h, k), replace x by x − h and y by y − k.

Horizontal opening

\[ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1. \]

• Vertices: (h − a, k), (h + a, k).
• Foci: (h − c, k), (h + c, k).

Asymptotes:

\[ y-k=\pm\frac ba(x-h). \]

Vertical opening

\[ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1. \]

• Vertices: (h, k − a), (h, k + a).
• Foci: (h, k − c), (h, k + c).

Asymptotes:

\[ y-k=\pm\frac ab(x-h). \]

Translation changes the position, but not a, b, c, or eccentricity.

Worked example 7: Reading a translated equation

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

\[ \frac{(y+1)^2}{9}-\frac{(x-2)^2}{16}=1. \]

The center is:

Center: (2, −1).

The positive term contains y, so the opening is vertical.

\[ a=3,\qquad b=4,\qquad c=5. \]

Move 3 units up and down from the center to find the vertices.

Vertices: (2, −4), (2, 2).

Move 5 units up and down to find the foci.

Foci: (2, −6), (2, 4).

Asymptotes:

\[ y+1=\pm\frac34(x-2). \]

Eccentricity:

\[ e=\frac53. \]

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15. Worked example 8: Completing the square

Convert to standard form and identify the center and asymptotes:

\[ 9x^2-4y^2-18x-16y-43=0. \]

Step 1: Group the terms

\[ 9(x^2-2x)-4(y^2+4y)=43. \]

Step 2: Complete the squares

\[ 9\big((x-1)^2-1\big) -4\big((y+2)^2-4\big)=43. \]

Expand the constant terms:

\[ 9(x-1)^2-9-4(y+2)^2+16=43. \]

Therefore:

\[ 9(x-1)^2-4(y+2)^2=36. \]

Step 3: Normalize

\[ \frac{(x-1)^2}{4}-\frac{(y+2)^2}{9}=1. \]

Center: (1, −2).

The opening is horizontal.

\[ a=2,\qquad b=3. \]

Asymptotes:

\[ y+2=\pm\frac32(x-1). \]

Sign warning: The factor −4 multiplies the entire completed-square expression. Distribute it carefully.

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16. Checking points and simple intersections

To check whether a point lies on a hyperbola, substitute its coordinates into the equation.

Worked example 9

Find the intersection points of x = 5 with:

\[ \frac{x^2}{9}-\frac{y^2}{16}=1. \]

Substitute:

\[ \frac{25}{9}-\frac{y^2}{16}=1. \]

Rearrange:

\[ \frac{y^2}{16}=\frac{16}{9}. \]

Therefore:

\[ y^2=\frac{256}{9}. \]\[ y=\pm\frac{16}{3}. \]

Intersection points: (5, 16/3), (5, −16/3).

What if x = 2?

Substituting x = 2 gives:

\[ \frac49-\frac{y^2}{16}=1. \]

Thus:

\[ y^2=-\frac{80}{9}. \]

There are no real solutions.

This agrees with the coordinate restriction: this horizontal hyperbola requires x ≤ −3 or x ≥ 3.

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17. Common mistakes and verification

Mistake 1: Using the ellipse relationship

For a hyperbola:

\[ c^2=a^2+b^2. \]

The focal distance is found by adding the squared parameters.

Mistake 2: Choosing a² as the larger denominator

For a hyperbola, a² is the denominator of the positive squared term in normalized form.

It does not have to be larger.

Mistake 3: Reading direction before normalizing

The opening rule applies when the right-hand side is positive 1.

Mistake 4: Reversing the asymptote slope

Use the correct formula or set the right-hand side of the normalized equation to zero and solve.

Mistake 5: Plotting conjugate-axis endpoints on the curve

They are construction points, not points on the hyperbola.

Mistake 6: Omitting absolute value in the definition

The definition uses the absolute difference of the focal distances.

Mistake 7: Treating the center as a point on the curve

At the center, both squared terms are zero, giving 0 rather than 1.

Mistake 8: Ignoring degeneracy

For example:

\[ \frac{x^2}{9}-\frac{y^2}{16}=0 \]

represents two intersecting lines, not a nondegenerate hyperbola.

Quick verification checklist

• Is the right-hand side positive 1?
• Are the squared terms subtracted?
• Does the positive term agree with the opening direction?
• Does c² = a² + b²?
• Is c > a?
• Is e > 1?
• Do both vertices satisfy the equation?
• Do the asymptotes intersect at the center?

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18. Independent practice

Complete these before reading the solutions.

Practice 1

For:

\[ \frac{x^2}{25}-\frac{y^2}{144}=1, \]

find the center, vertices, foci, axis lengths, asymptotes, and eccentricity.

Practice 2

For:

\[ \frac{y^2}{16}-\frac{x^2}{9}=1, \]

find the vertices, foci, asymptotes, and eccentricity.

Practice 3

A hyperbola has vertices (−5, 0), (5, 0) and foci (−√34, 0), (√34, 0).

Find its equation.

Practice 4

A hyperbola is centered at the origin and opens upward and downward. Its vertices are (0, −4), (0, 4), and its asymptotes are:

\[ y=\pm2x. \]

Find its equation.

Practice 5

A horizontal hyperbola centered at the origin has a = 6 and eccentricity 5/3.

Find its equation.

Practice 6

Convert to standard form and find the foci:

\[ 9x^2-16y^2=-144. \]

Practice 7

Find the center, vertices, and asymptotes of:

\[ \frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1. \]

Practice 8

The constant absolute focal-distance difference is 10. For a particular point, the greater focal distance is 18.

Find the smaller distance.

Practice 9

Find all intersection points of y = 3 with:

\[ \frac{x^2}{8}-\frac{y^2}{9}=1. \]

Practice 10

Convert to standard form and identify the center:

\[ 4x^2-9y^2-16x-18y-29=0. \]

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19. Practice solutions

Solution 1

\[ a=5,\qquad b=12,\qquad c=\sqrt{25+144}=13. \]

• Center: (0, 0).
• Vertices: (−5, 0), (5, 0).
• Foci: (−13, 0), (13, 0).
• Transverse axis length: 10.
• Conjugate axis length: 24.
• Eccentricity: 13/5.

Asymptotes:

\[ y=\pm\frac{12}{5}x. \]

Solution 2

The opening is vertical.

\[ a=4,\qquad b=3,\qquad c=5. \]

Vertices: (0, −4), (0, 4).

Foci: (0, −5), (0, 5).

Eccentricity: 5/4.

Asymptotes:

\[ y=\pm\frac43x. \]

Solution 3

\[ a=5,\qquad c=\sqrt{34}. \]\[ b^2=34-25=9. \]

Therefore:

\[ \frac{x^2}{25}-\frac{y^2}{9}=1. \]

Solution 4

The vertical vertices give a = 4.

For a vertical hyperbola:

\[ \frac ab=2. \]

Thus b = 2.

The equation is:

\[ \frac{y^2}{16}-\frac{x^2}{4}=1. \]

Solution 5

\[ c=ae=6\left(\frac53\right)=10. \]\[ b^2=100-36=64. \]

Therefore:

\[ \frac{x^2}{36}-\frac{y^2}{64}=1. \]

Solution 6

Rewrite and divide by 144:

\[ 16y^2-9x^2=144. \]\[ \frac{y^2}{9}-\frac{x^2}{16}=1. \]\[ a=3,\qquad b=4,\qquad c=5. \]

Foci: (0, −5), (0, 5).

Solution 7

Center: (−1, 2).

The opening is horizontal and a = 3.

Vertices: (−4, 2), (2, 2).

Asymptotes:

\[ y-2=\pm\frac43(x+1). \]

Solution 8

Let the smaller distance be d.

\[ 18-d=10. \]

Smaller distance: 8 units.

Solution 9

Substitute y = 3:

\[ \frac{x^2}{8}-1=1. \]\[ x^2=16. \]

Intersection points: (−4, 3), (4, 3).

Solution 10

Group:

\[ 4(x^2-4x)-9(y^2+2y)=29. \]

Complete the squares:

\[ 4\big((x-2)^2-4\big) -9\big((y+1)^2-1\big)=29. \]

Therefore:

\[ 4(x-2)^2-9(y+1)^2=36. \]

Standard form:

\[ \frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1. \]

Center: (2, −1).

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20. Lesson assessment — 03.5 Hyperbolas Quiz

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

Question 1

Which equation represents a hyperbola opening upward and downward?

A. \(x^2/16-y^2/9=1\)

B. \(y^2/9-x^2/16=1\)

C. \(x^2/16+y^2/9=1\)

D. \(y^2=4x\)

Question 2

Find the foci of:

\[ \frac{x^2}{16}-\frac{y^2}{9}=1. \]

A. (−√7, 0), (√7, 0)

B. (0, −5), (0, 5)

C. (−4, 0), (4, 0)

D. (−5, 0), (5, 0)

Question 3

Find the asymptotes of:

\[ \frac{x^2}{9}-\frac{y^2}{25}=1. \]

A. \(y=\pm\frac53x\)

B. \(y=\pm\frac35x\)

C. \(y=\pm\frac{25}{9}x\)

D. \(y=\pm\frac95x\)

Question 4

A hyperbola has a = 8 and b = 6.

What is its eccentricity?

A. 3/4

B. 4/5

C. 5/4

D. 5/3

Question 5

A hyperbola has vertices (0, −3), (0, 3) and foci (0, −5), (0, 5).

Which is its equation?

A. \(x^2/9-y^2/16=1\)

B. \(y^2/9-x^2/16=1\)

C. \(y^2/25-x^2/9=1\)

D. \(y^2/9+x^2/16=1\)

Question 6

For:

\[ \frac{x^2}{25}-\frac{y^2}{11}=1, \]

the greater focal distance of a point on the curve is 14.

What is its smaller focal distance?

A. 9

B. 10

C. 24

D. 4

Question 7

Find the center of:

\[ \frac{(x+3)^2}{4}-\frac{(y-1)^2}{9}=1. \]

A. (−3, 1)

B. (3, −1)

C. (−3, −1)

D. (3, 1)

Question 8

Find the vertices of:

\[ 4x^2-9y^2=-36. \]

A. (−3, 0), (3, 0)

B. (−2, 0), (2, 0)

C. (0, −2), (0, 2)

D. (0, −3), (0, 3)

Question 9

A horizontal hyperbola centered at the origin has a = 3 and eccentricity 5/3.

Which is its equation?

A. \(x^2/9-y^2/25=1\)

B. \(x^2/9-y^2/16=1\)

C. \(y^2/9-x^2/16=1\)

D. \(x^2/16-y^2/9=1\)

Question 10

Find the intersection points of y = 3 with:

\[ \frac{x^2}{8}-\frac{y^2}{9}=1. \]

A. (√8, 3), (−√8, 3)

B. (2, 3), (−2, 3)

C. There are no real intersection points.

D. (4, 3), (−4, 3)

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21. Instructor answer key and assessment solutions

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

Question	Answer	Skill assessed
1	B	Opening direction
2	D	Foci
3	A	Asymptotes
4	C	Eccentricity
5	B	Equation from vertices and foci
6	D	Focal-distance difference
7	A	Translated center
8	C	Normalization and vertices
9	B	Equation from eccentricity
10	D	Simple intersections

Answer 1: B

In normalized form, the positive y² term indicates vertical opening.

Answer 2: D

\[ c=\sqrt{16+9}=5. \]

The transverse axis is horizontal.

Foci: (−5, 0), (5, 0).

Answer 3: A

\[ a=3,\qquad b=5. \]

For a horizontal hyperbola:

\[ y=\pm\frac ba x=\pm\frac53x. \]

Answer 4: C

\[ c=\sqrt{8^2+6^2}=\sqrt{100}=10. \]\[ e=\frac{10}{8}=\frac54. \]

Answer 5: B

The vertices give a = 3, and the foci give c = 5.

\[ b^2=25-9=16. \]

The transverse axis is vertical:

\[ \frac{y^2}{9}-\frac{x^2}{16}=1. \]

Answer 6: D

\[ a=5, \]

so the constant difference is 10.

\[ 14-d=10. \]

Smaller distance: 4 units.

Answer 7: A

Read the squared expressions as x − h and y − k.

Center: (−3, 1).

Answer 8: C

Normalize:

\[ \frac{y^2}{4}-\frac{x^2}{9}=1. \]

The opening is vertical and a = 2.

Vertices: (0, −2), (0, 2).

Answer 9: B

\[ c=ae=3\left(\frac53\right)=5. \]\[ b^2=25-9=16. \]

Therefore:

\[ \frac{x^2}{9}-\frac{y^2}{16}=1. \]

Answer 10: D

Substitute y = 3:

\[ \frac{x^2}{8}-1=1. \]\[ x^2=16. \]

Intersection points: (4, 3), (−4, 3).

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22. Revision guidance

• Missed Questions 1 or 8: Review normalization and opening direction.
• Missed Question 2: Review c² = a² + b².
• Missed Question 3: Review asymptote slopes and the construction rectangle.
• Missed Questions 4 or 9: Review eccentricity.
• Missed Question 5: Review constructing equations from vertices and foci.
• Missed Question 6: Review the absolute focal-distance difference.
• Missed Question 7: Review translated centers and signs.
• Missed Question 10: Review substitution and both square-root signs.

For each incorrect or guessed answer:

1. Identify the mistake.
2. Write a corrected solution.
3. Verify the result using substitution or another geometric property.

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23. Lesson review

You should now be able to:

• Define a hyperbola using its two foci.
• Distinguish the transverse and conjugate axes.
• Recognize horizontal and vertical standard equations.
• Identify centers, vertices, and foci.
• Find and use asymptotes.
• Calculate eccentricity.
• Construct equations from vertices, foci, asymptotes, or eccentricity.
• Read translated equations and complete the square.
• Check points and solve simple intersections.
• Distinguish hyperbola properties from ellipse properties.

The central relationships are:

\[ c^2=a^2+b^2, \]\[ e=\frac ca>1, \]

and:

\[ |PF_1-PF_2|=2a. \]

Completion target: Score at least 8/10 on the lesson quiz and independently correct every missed or guessed answer.