The difference-of-distances definition of a hyperbola, the two standard equations with centre at the origin, the central rectangle and the asymptotes, identifying the transverse axis and locating vertices and foci, writing an equation from foci and vertices, and modelling a hyperbolic mirror.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Graph and Write Equations of Hyperbolas
Objectives
Five outcomes. Change one word in the ellipse's definition and the curve breaks open.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-647 — the lesson these objectives are drawn from
Warm-up
Lesson 9.4 built an ellipse by fixing the SUM of the distances to two foci.
Discussion prompt
Suppose instead you fix the DIFFERENCE of the two distances at 6, with the foci 10 apart. Can a point be found? What if the difference were 12?
Hint: Compare the difference with the distance between the foci.
Answer:
A difference of 6 is possible: the point 3 units from the centre along the line of the foci is 2 from one and 8 from the other.
\[ |d_2-d_1| = 6 < 10 \quad \checkmark; \qquad |d_2-d_1| = 12 > 10 \quad \times \]
A difference of 12 is impossible, since no side of a triangle can exceed the sum of the other two. The possible points form a hyperbola, with two separate branches — one for each sign of the difference.
Concept
A hyperbola is the set of points for which the distances to two fixed foci differ by a constant. That constant is the distance between the two vertices, and the curve has two branches and two asymptotes.
hyperbola — The set of all points in a plane for which the difference of the distances to two fixed points, the foci, is constant. The transverse axis joins the two vertices and its midpoint is the centre.
\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad c^2 = a^2+b^2 \]
Because the foci lie beyond the vertices here, c is larger than a — so the relation between the three carries a plus rather than the ellipse's minus.
Figure (svg): Two columns comparing an ellipse with a hyperbola
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-642
Section
Section 1
Concept
The line through the two foci meets the hyperbola at the two vertices, joined by the transverse axis whose midpoint is the centre. The constant difference equals the length of that axis.
\[ |d_2-d_1| = 2a \]
The two branches come from the two signs of the difference: one branch is nearer the first focus, the other nearer the second.
Figure (svg): A hyperbola with two foci and a point on it, showing the two distances whose difference is constant
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-642 — Definition of a hyperbola
Picture it
A hyperbola with two of its points and both pairs of distances.
Figure (svg): A hyperbola with two foci and a point on it, showing the two distances whose difference is constant
At the vertex the distances are 2 and 8; higher up they are 5.33 and 11.33. Both differ by 6, which is the distance between the two vertices.
Worked example
Verifying the definition at two points of a hyperbola with foci 5 units out.
\[ \text{For foci } (\pm 5, 0) \text{ and } a = 3, \text{ check the vertex } (3,0) \text{ and the point } (5, 5.33). \]
Vertex: distances to the two foci
Why: Two to the near one, 8 to the far one.
\[ 2\text{ and } 8 \]
Vertex: subtract
Why: Eight minus 2.
\[ 6 \]
Second point: distances
Why: Straight up 5.33 from one focus, and 11.33 by the distance formula from the other.
\[ 5.33\text{ and } 11.33 \]
Second point: subtract
Why: Eleven point three three minus 5.33.
\[ 6 \]
Figure (svg): A hyperbola with two foci and a point on it, showing the two distances whose difference is constant
\[ |d_2-d_1| = 6 = 2a \]
Verify: explain why the difference equals 2a
Why: At a vertex the distances are c minus a and c plus a, and their difference is 2a whatever c is. So the constant is always the distance between the two vertices — exactly parallel to the ellipse, where the constant sum was the distance between them there too.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-642
Comparison
Fill the blanks. One word, then one sign, then one ordering.
Comparison matrix
| Question | Ellipse | Hyperbola |
|---|---|---|
| The definition fixes | the sum of the distances | the difference of the distances |
| Sign between the squares | plus | minus |
| Relation for c | c^2 = a^2 - b^2 | c^2 = a^2 + b^2 |
| Which is largest | a | c |
Every row differs by exactly one thing, and the four differences are all consequences of the first: fixing a difference rather than a sum.
Worked example
Reading the labelled diagrams of the key concept.
\[ \text{For } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text{ name the centre, vertices, foci and asymptotes.} \]
The centre
Why: The midpoint of the transverse axis.
\[ (0, 0) \]
The vertices
Why: On the horizontal axis, a units out.
\[ (a, 0)\text{ and } (-a, 0) \]
The foci
Why: Farther out on the same axis, c units.
\[ (c, 0)\text{ and } (-c, 0) \]
The asymptotes
Why: Diagonals of the rectangle 2a wide by 2b tall.
\[ y = +- (\frac{b}{a}) x \]
Figure (svg): The parts of a hyperbola in both orientations, with the central rectangle and asymptotes
\[ (\pm a,0), \; (\pm c,0), \; y = \pm\tfrac{b}{a}x \]
Verify: check that c exceeds a
Why: Since c squared equals a squared plus b squared and b is positive, c is greater than a — so the foci lie beyond the vertices, outside the region between the branches. That is the opposite of the ellipse, where the foci sat inside, and it is worth holding onto as the distinguishing fact.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-642
Trap
\[ \text{foci } (\pm 4,0), \text{ vertices } (\pm 3,0) \]
Assume the vertices must be farther out
Why: The ellipse's arrangement, with foci inside, is carried over.
\[ \text{swap them: } a = 4, \; c = 3 \quad \text{(wrong)} \]
Then b squared would be 9 minus 16, a negative number, and no real hyperbola exists.
\[ a = 3, \; c = 4 \;\Longrightarrow\; b^2 = 16-9 = 7 \]
For a hyperbola the FOCI are farther out
Why: The curve bends away from the centre, so it reaches the vertices before the foci.
\[ c > a \text{ always, for a hyperbola} \]
The relation itself encodes the ordering: c squared equals a squared plus b squared makes c the largest of the three, whereas for an ellipse a was.
Fill the middle
At a vertex of a hyperbola.
Fill in the blanks
(c+a)-(c-a) = 2a
Why: The c terms cancel, leaving 2a — the distance between the two vertices. So the constant in the definition is always the transverse axis's full length.
Sorting
The foci lie beyond the vertices.
Sort into buckets
Sort each statement.
The last row is a real difference from the ellipse, where a was always the larger. For a hyperbola, a is simply whichever denominator sits under the positive term.
Prediction
Commit before reasoning.
Predict first
Why does a hyperbola have two separate branches while an ellipse has one closed curve?
Correct: Because the difference can be positive or negative, giving one branch nearer each focus.
\[ d_2-d_1 = 2a \text{ or } d_1-d_2 = 2a \]
Why: The definition uses the absolute difference, so points nearer the left focus by 2a form one branch and points nearer the right focus by 2a form the other. An ellipse's sum is a single non-negative quantity with no such choice, so it yields one curve. The two branches are not two separate curves that happen to share an equation — they are the two cases of one condition.
Section
Section 2
Concept
With centre at the origin the equation is x squared over a squared minus y squared over b squared equal to 1 for a horizontal transverse axis, and the two terms exchanged for a vertical one. The asymptotes are the diagonals of the rectangle 2a by 2b.
\[ y = \pm\frac{b}{a}x \quad \text{or} \quad y = \pm\frac{a}{b}x \]
It is the sign, not the size, that decides here: a belongs to whichever variable carries the positive term, whether or not its denominator is the larger.
Figure (svg): The parts of a hyperbola in both orientations, with the central rectangle and asymptotes
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-642 — Standard Equation of a Hyperbola with Center at the Origin
Picture it
The standard hyperbolas with their central rectangles and asymptotes.
Figure (svg): The parts of a hyperbola in both orientations, with the central rectangle and asymptotes
The rectangle's corners lie on the asymptotes and its sides touch the vertices. Drawing it first makes the branches almost draw themselves.
Worked example
Applying the rule to Example 1 and the guided practice.
\[ \text{Which axis is transverse for } \tfrac{y^2}{4}-\tfrac{x^2}{25}=1, \; \tfrac{x^2}{16}-\tfrac{y^2}{49}=1, \; \tfrac{y^2}{36}-x^2=1? \]
First: which term is positive
Why: The y squared term.
Second: which term is positive
Why: The x squared term.
Third: which term is positive
Why: The y squared term, with an invisible denominator of 1 under x squared.
Note what does not matter
Why: In the first, b exceeds a, and that is allowed.
Figure (svg): The parts of a hyperbola in both orientations, with the central rectangle and asymptotes
\[ V, \; H, \; V \]
Verify: contrast with the ellipse rule
Why: For an ellipse the LARGER denominator marked the major axis; for a hyperbola the POSITIVE term marks the transverse axis, whatever the sizes. In the first example a is 2 and b is 5, so the larger denominator sits under the negative term — a combination no ellipse could have.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-644
Sorting
Find the positive term.
Sort into buckets
Sort each hyperbola.
In the first item the larger denominator sits under the negative term, which would be impossible for an ellipse — a reminder that the two families use different rules.
Worked example
Example 1 Step 2 and the guided practice.
\[ \text{Find the asymptotes of } \tfrac{y^2}{4}-\tfrac{x^2}{25}=1 \text{ and } \tfrac{x^2}{16}-\tfrac{y^2}{49}=1. \]
First: read a and b
Why: Two and 5, with a from the positive term.
\[ a = 2, b = 5 \]
First: vertical transverse axis
Why: The slopes are plus and minus a over b.
\[ y = +- (\frac{2}{5}) x \]
Second: read a and b
Why: Four and 7, with a from the positive term.
\[ a = 4, b = 7 \]
Second: horizontal transverse axis
Why: The slopes are plus and minus b over a.
\[ y = +- (\frac{7}{4}) x \]
Figure (svg): The solution to Worked example find the asymptotes shown as a ladder of expressions, one row per algebraic move
\[ y = \pm\tfrac{2}{5}x; \quad y = \pm\tfrac{7}{4}x \]
Verify: read the slopes off the rectangle
Why: The rectangle for the first is 4 tall and 10 wide, so its diagonal rises 2 for every 5 across — a slope of two fifths, matching. Reading the slope as rise over run on the rectangle avoids memorising which of a over b and b over a applies, since the picture gives it directly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-643
Error analysis
A student identifies the transverse axis of a hyperbola.
Annotate
On: \( \frac{y^2}{4}-\frac{x^2}{25}=1 \;\Longrightarrow\; \text{transverse axis horizontal, since } 25 > 4 \)
Sign for a hyperbola, size for an ellipse. Mixing the two rules reverses the orientation and every part that depends on it.
Fill the middle
Example 1.
Fill in the blanks
\tfrac5___-\tfrac______=1: \; y = \pm\frac______}x
Why: With a vertical transverse axis the slopes are plus and minus a over b, which is two fifths. Reading them off the rectangle — 2 up for 5 across — gives the same answer without memorising the formula.
Matching
Rise over run on the central rectangle.
Match the pairs
Why: The third has an invisible denominator of 1 under x squared, so b is 1 and the asymptotes are very steep. A wide rectangle gives shallow asymptotes and a tall narrow one gives steep asymptotes.
Prediction
Commit before reasoning.
Predict first
Why does the curve x squared over 9 minus y squared over 7 equal 1 flatten toward its asymptotes far out?
Correct: Because for huge x the 1 becomes negligible, leaving y approximately equal to plus or minus (b/a)x.
\[ y^2 = b^2\left(\frac{x^2}{a^2}-1\right) \approx \frac{b^2x^2}{a^2} \text{ for } |x| \text{ large} \]
Why: Rearranging gives y squared equal to b squared times x squared over a squared minus 1, and for large x the 1 is negligible beside the first term. So y is approximately plus or minus b over a times x — the asymptote equations. This is the same reasoning that gave rational functions their horizontal asymptotes in Lesson 8.3: far from the origin, the smaller terms stop mattering.
Section
Section 3
Concept
Divide into standard form, read a from the positive term and b from the negative one, draw the rectangle 2a by 2b centred at the origin, extend its diagonals as asymptotes, and sketch the branches through the vertices.
\[ c^2 = a^2+b^2 \]
The rectangle does three jobs at once: its edge midpoints are the vertices, its diagonals are the asymptotes, and its proportions show how widely the branches open.
Figure (svg): A hyperbola with a vertical transverse axis, drawn using its central rectangle and asymptotes
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-643 — Graph an equation of a hyperbola
Picture it
Example 1, with the rectangle and asymptotes drawn.
Figure (svg): A hyperbola with a vertical transverse axis, drawn using its central rectangle and asymptotes
The rectangle is 4 tall and 10 wide, so the asymptotes are shallow. The branches leave the vertices at 0 comma 2 and 0 comma negative 2 and hug those lines.
Worked example
Example 1.
\[ \text{Graph } 25y^2-4x^2=100 \text{ and identify the vertices, foci and asymptotes.} \]
Divide into standard form
Why: Every term over 100.
\[ y ^{2} / 4 - x ^{2} / 25 = 1 \]
Read a and b
Why: The square roots of 4 and 25, with a from the positive term.
\[ a = 2, b = 5 \]
State vertices and asymptotes
Why: Vertical transverse axis, so vertices on the vertical axis.
\[ (0, +- 2); y = +- (\frac{2}{5}) x \]
Find c
Why: Four plus 25.
\[ c = \sqrt{29}\text{ about } 5.4 \]
Figure (svg): A hyperbola with a vertical transverse axis, drawn using its central rectangle and asymptotes
\[ (0,\pm 2), \; (0,\pm\sqrt{29}), \; y = \pm\tfrac{2}{5}x \]
Verify: check that c exceeds a
Why: The square root of 29 is about 5.39, comfortably more than a equal to 2 — so the foci lie beyond the vertices, outside the region between the branches. For an ellipse the reverse would hold, and getting c smaller than a here would signal that the plus had been written as a minus.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-643
Fill the middle
Example 1.
Fill in the blanks
c^2 = 4+25 = 29
Why: Four plus 25 is 29, so c is about 5.39 — larger than a, which places the foci beyond the vertices as a hyperbola requires.
Worked example
Guided Practice 1 to 3.
\[ \text{Identify the parts of } \tfrac{x^2}{16}-\tfrac{y^2}{49}=1, \; \tfrac{y^2}{36}-x^2=1, \; 4y^2-9x^2=36. \]
First: horizontal, a is 4 and b is 7
Why: Sixteen plus 49 is 65.
\[ (+- 4, 0); (+- \sqrt{65}, 0); y = +- (\frac{7}{4}) x \]
Second: vertical, a is 6 and b is 1
Why: Thirty-six plus 1 is 37.
\[ (0, +- 6); (0, +- \sqrt{37}); y = +- 6 x \]
Third: divide by 36
Why: Y squared over 9 minus x squared over 4.
Third: find c and the asymptotes
Why: Nine plus 4 is 13.
\[ (0, +- \sqrt{13}); y = +- (\frac{3}{2}) x \]
Figure (svg): The solution to Worked example three more hyperbolas shown as a ladder of expressions, one row per algebraic move
\[ (\pm\sqrt{65},0); \; (0,\pm\sqrt{37}); \; (0,\pm\sqrt{13}) \]
Verify: check the second's steep asymptotes
Why: With b equal to 1 the rectangle is 12 tall and only 2 wide, so its diagonals have slope 6 — very steep, and the branches barely spread sideways. A tall narrow rectangle always means a narrow hyperbola, which is a useful shape check before plotting anything.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 644-644
Trap
\[ \frac{y^2}{4}-\frac{x^2}{25}=1 \]
Use the ellipse's relation
Why: The formula from the previous lesson is applied unchanged.
\[ c^2 = 4-25 = -21 \quad \text{(no real } c\text{)} \]
A negative square leaves no real foci, which cannot be right — every hyperbola has two.
\[ c^2 = a^2+b^2 = 4+25 = 29 \]
For a hyperbola the relation carries a PLUS
Why: The foci lie beyond the vertices, so c must exceed a.
\[ c = \sqrt{29} \approx 5.4 > 2 \quad \checkmark \]
The sign matches the equation's: minus between the squares, plus in the relation for c. Ellipses have it the other way round in both places.
Ranking
Graphing a hyperbola from any equation.
Put in order
Why: The rectangle in step three is doing the real work: without it the asymptotes have to be computed from a formula and the branches drawn by guesswork. With it, everything after is a matter of tracing.
Sorting
Ellipse or hyperbola.
Sort into buckets
Sort each situation by the relation it needs.
The sign in the relation always matches the geometry, so remembering where the foci sit is enough to reconstruct the formula.
Prediction
Commit before reasoning.
Predict first
Example 1's rectangle is 4 tall and 10 wide. What does that say about the branches?
Correct: They open widely, since the asymptotes have shallow slopes of plus and minus two fifths.
\[ \text{slope} = \frac{\text{half-height}}{\text{half-width}} = \frac{2}{5} \]
Why: The rectangle's proportions are the asymptote slopes, and a shallow slope means the branches spread sideways quickly as they rise. Compare Guided Practice 2, where the rectangle is 12 tall and 2 wide: slopes of plus and minus 6, and branches that shoot almost straight up. Sketching the rectangle first therefore tells you the shape of the answer before any point is plotted.
Section
Section 4
Concept
The vertices and foci both lie on the transverse axis, so their positions fix the orientation. The vertices give a, the foci give c, and b squared comes from c squared minus a squared.
\[ b^2 = c^2-a^2 \]
Note the rearrangement: the relation is c squared equals a squared plus b squared, so isolating b squared subtracts. For an ellipse the same isolation gave b squared equal to a squared minus c squared.
Figure (svg): A hyperbola built from its two foci and two vertices
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-643 — Write an equation of a hyperbola
Picture it
Example 2: two foci and two vertices, all on the horizontal axis.
Figure (svg): A hyperbola built from its two foci and two vertices
The vertices at 3 give a, the foci at 4 give c, and b squared is 16 minus 9, or 7. The equation follows immediately.
Worked example
Example 2.
\[ \text{Write the hyperbola with foci } (\pm 4, 0) \text{ and vertices } (\pm 3, 0). \]
Read the orientation
Why: All four points lie on the horizontal axis.
Read a and c
Why: Three from the vertices, 4 from the foci.
\[ a = 3, c = 4 \]
Find b squared
Why: Sixteen minus 9.
\[ b ^{2} = 7 \]
Substitute into the form
Why: Nine under x squared, 7 under y squared, with a minus.
\[ x ^{2} / 9 - y ^{2} / 7 = 1 \]
Figure (svg): A hyperbola built from its two foci and two vertices
\[ \frac{x^2}{9}-\frac{y^2}{7}=1 \]
Verify: test a vertex
Why: At 3 comma 0 the equation gives 9 over 9 minus 0, which is 1 — correct. And c is 4, greater than a equal to 3, so the foci really do lie beyond the vertices. Both checks together confirm the orientation and the subtraction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 643-643
Fill the middle
Example 2.
Fill in the blanks
b^2 = c^2-a^2 = 16-9 = 7
Why: Sixteen minus 9 is 7, so the denominator under y squared is 7. Note the subtraction runs c squared minus a squared here, the reverse of the ellipse's a squared minus c squared.
Worked example
Guided Practice 4 and 5.
\[ \text{Write the hyperbolas with foci } (\pm 3,0), \text{ vertices } (\pm 1,0); \text{ and foci } (0,\pm 10), \text{ vertices } (0,\pm 6). \]
First: horizontal, a is 1 and c is 3
Why: Nine minus 1.
\[ b ^{2} = 8 \]
First: write the equation
Why: One under x squared and 8 under y squared.
\[ x ^{2} - y ^{2} / 8 = 1 \]
Second: vertical, a is 6 and c is 10
Why: One hundred minus 36.
\[ b ^{2} = 64 \]
Second: write the equation
Why: Thirty-six under y squared and 64 under x squared.
\[ y ^{2} / 36 - x ^{2} / 64 = 1 \]
Figure (svg): The solution to Worked example two more equations shown as a ladder of expressions, one row per algebraic move
\[ x^2-\tfrac{y^2}{8}=1; \quad \tfrac{y^2}{36}-\tfrac{x^2}{64}=1 \]
Verify: check the second's asymptotes for plausibility
Why: With a equal to 6 and b equal to 8 the rectangle is 12 by 16, so the asymptotes have slope three quarters — a wide-opening hyperbola. And 10 is greater than 6, so the foci lie beyond the vertices. In the first, b squared came out larger than a squared, which is perfectly allowed here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 644-644
Error analysis
A student writes a hyperbola with vertices at 3 and foci at 4.
Annotate
On: \( b^2 = a^2-c^2 = 9-16 = -7 \)
A negative value for b squared is an immediate signal that the wrong relation was used, since b squared is a square and cannot be negative.
Matching
Vertices give a; foci give c.
Match the pairs
Why: In every case the vertices' axis is the transverse one and the foci lie farther out along it. The last is Example 1 read backwards, which is a useful way to check a graphing answer.
Two truths and a lie
All three are about writing a hyperbola's equation.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Example 1 has a equal to 2 and b equal to 5, and Guided Practice 4 gives a equal to 1 with b squared equal to 8. For a hyperbola a is defined by the SIGN of its term, not by being the larger — which is exactly the rule that differs from the ellipse.
Prediction
Commit before reasoning.
Predict first
Someone claims a hyperbola has foci at plus and minus 3 and vertices at plus and minus 4. What goes wrong?
Correct: b squared would be 9 minus 16, which is negative, so no such hyperbola exists.
\[ c > a: \text{ hyperbola}; \qquad c < a: \text{ ellipse} \]
Why: For a hyperbola the foci must lie beyond the vertices, so c must exceed a — and here it does not. The algebra reports the impossibility as a negative value for a square. Those numbers would describe a perfectly good ELLIPSE, with a equal to 4 and c equal to 3, which is how the two families divide up the possible arrangements between them.
Section
Section 5
Concept
A hyperbolic mirror reflects rays aimed at one focus toward the other. Placing a camera at the second focus captures a full circular view, which software can then unwarp.
\[ \frac{y^2}{7.90}-\frac{x^2}{5.50}=1 \]
Only one branch is the physical mirror. The equation produces both, and reading which one the situation uses is part of the modelling.
Figure (svg): The cross section of a hyperbolic panoramic mirror, with its two foci and its width
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 644-644 — Hyperbolic mirror
Picture it
Example 3: a mirror with a equal to 2.81 and c equal to 3.66 centimetres.
Figure (svg): The cross section of a hyperbolic panoramic mirror, with its two foci and its width
The solid upper branch is the mirror; the dashed lower one is the equation's other half. At 3 centimetres out the mirror is at height 4.56.
Worked example
Example 3.
\[ \text{A mirror has } a = 2.81 \text{ and } c = 3.66 \text{ cm. Write its cross section and find its height if it is } 6 \text{ cm wide.} \]
Find b squared
Why: Three point six six squared minus 2.81 squared.
\[ \text{about } 5.50 \]
Write the equation
Why: Vertical transverse axis, a squared about 7.90.
\[ y ^{2} / 7.90 - x ^{2} / 5.50 = 1 \]
Substitute the half-width
Why: Six centimetres wide, so x is 3.
\[ y ^{2} / 7.90 - \frac{9}{5.50} = 1 \]
Solve for y
Why: Y squared is 7.90 times 2.636.
\[ y\text{ about } 4.56 \]
Figure (svg): The cross section of a hyperbolic panoramic mirror, with its two foci and its width
\[ y \approx 4.56, \quad \text{height } 4.56-2.81 \approx 1.75 \]
Verify: measure from the vertex
Why: The vertex sits at 2.81 and the rim at 4.56, so the mirror's depth is the difference, about 1.75 centimetres. Subtracting the vertex height is essential: the y-coordinate 4.56 measures from the centre of the hyperbola, which is not on the mirror at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 644-644
Fill the middle
Example 3, Step 1.
Fill in the blanks
b^2 = 3.66^2-2.81^2 \approx 5.50
Why: Thirteen point four minus 7.90 is about 5.50, the denominator under x squared. The subtraction runs c squared minus a squared, as it must for a hyperbola.
Worked example
The same model, extended.
\[ \text{How tall would the same mirror be if it were } 8 \text{ cm wide instead of } 6? \]
Use the same equation
Why: Only the width changes.
\[ y ^{2} / 7.90 - x ^{2} / 5.50 = 1 \]
Substitute the new half-width
Why: Eight centimetres wide, so x is 4.
\[ y ^{2} / 7.90 = 1 + \frac{16}{5.50} \]
Compute
Why: One plus 2.909 is 3.909, times 7.90.
\[ y ^{2}\text{ about } 30.9 \]
Take the root and subtract
Why: Five point five six minus 2.81.
\[ \text{height about } 2.75 \text{cm} \]
Figure (svg): The solution to Worked example a wider mirror shown as a ladder of expressions, one row per algebraic move
\[ y \approx 5.56, \quad \text{height} \approx 2.75 \text{ cm} \]
Verify: compare the two widths
Why: Going from 6 to 8 centimetres wide raises the height from 1.75 to 2.75, so a third more width costs well over half again as much height. The branches steepen as they rise toward the asymptotes, so the mirror gets deeper faster than it gets wider — which is why panoramic mirrors are made shallow.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 644-644
Trap
\[ y \approx 4.56 \text{ at the rim} \]
Report 4.56 as the mirror's height
Why: The computed coordinate is taken as the answer.
\[ \text{height} = 4.56 \text{ cm} \quad \text{(wrong)} \]
That measures from the CENTRE of the hyperbola, which lies below the mirror entirely — 2.81 centimetres below its lowest point.
\[ \text{height} = 4.56-2.81 \approx 1.75 \text{ cm} \]
Measure from the vertex, the mirror's own lowest point
Why: Coordinates are measured from the centre; the object starts at its vertex.
\[ \text{vertex at } y = a = 2.81 \]
The centre of a hyperbola is not on the curve at all, which is what makes this subtraction easy to forget and important to remember.
Prediction
Commit before reasoning.
Predict first
A parabolic mirror sends parallel rays to one focus. What does a hyperbolic mirror do instead?
Correct: It sends rays aimed at one focus toward the other, which is what puts the camera out of the way.
\[ \text{toward focus 1} \;\Longrightarrow\; \text{reflected to focus 2} \]
Why: The two-focus reflection property lets the camera sit at the second focus rather than in the middle of the incoming light, so it does not block the view it is recording. A parabola has only one focus, so a camera there would sit directly in the light path. Each conic's reflection property follows from its defining distance condition, which is why the shapes are chosen rather than approximated.
Comparison
Fill the blanks. The same curve, sampled twice.
Comparison matrix
| Quantity | 6 cm wide | 8 cm wide |
|---|---|---|
| Half-width x | 3 | 4 |
| y at the rim | about 4.56 | about 5.56 |
| Height above the vertex | about 1.75 cm | about 2.75 cm |
| Shape | shallow | noticeably deeper |
A third more width costs well over half again as much depth, because the branches steepen toward their asymptotes as they rise.
Ranking
From a mirror's dimensions to its height.
Put in order
Why: Step five is the one with no counterpart in the parabola problems of Lesson 9.2, where the vertex sat at the origin and the coordinate WAS the depth. Here the centre is not on the curve, so a subtraction is needed at the end.
Comparison
Fill the blanks. Each is a condition on distances.
Comparison matrix
| Curve | Distance condition | Equation at the origin |
|---|---|---|
| Parabola | equal to a point and a line | x^2 = 4py |
| Circle | fixed distance from one point | x^2 + y^2 = r^2 |
| Ellipse | fixed SUM of distances to two points | x^2/a^2 + y^2/b^2 = 1 |
| Hyperbola | fixed DIFFERENCE of distances to two points | x^2/a^2 - y^2/b^2 = 1 |
The last two differ by one word and one sign, and everything else — the ordering of a and c, the closed or open shape, the presence of asymptotes — follows from that.
Pattern
One routine to graph, one to write, one to model.
For a hyperbola the sign decides the orientation and c is the largest of the three. For an ellipse the size decides and a is the largest.
OpenStax Algebra and Trigonometry 2e, §12.2 The Hyperbola §12.2
Check
The sign decides, not the size.
Check your understanding
What are the vertices of 25y^2 - 4x^2 = 100?
Answer: A
Why: In standard form y^2/4 - x^2/25 = 1, the y term is positive, so a = 2 vertically.
Check
Foci. Add, do not subtract.
Check your understanding
Where are the foci of y^2/4 - x^2/25 = 1?
Answer: A
Why: c^2 = a^2 + b^2 = 4 + 25 = 29, on the vertical transverse axis.
Check
Vertices give a; foci give c.
Check your understanding
Write the hyperbola with foci (+-4, 0) and vertices (+-3, 0).
Answer: A
Why: With a = 3 and c = 4, b squared is 16 - 9, which is 7.
Real world
A long-range navigation system sends a pulse from two transmitters 400 kilometres apart. A ship measures the two arrival times and finds the pulse from the nearer station arrives earlier by an interval corresponding to a distance difference of 240 kilometres.
Discussion prompt
Taking the midpoint between the stations as the origin with them on the horizontal axis, find the hyperbola on which the ship lies, and explain why a second pair of stations is needed.
Hint: The constant difference is 2a, and the stations are the foci.
Answer:
\[ 2a = 240 \;\Longrightarrow\; a = 120; \qquad c = 200 \]
\[ b^2 = c^2-a^2 = 40{,}000-14{,}400 = 25{,}600 \]
\[ \frac{x^2}{14{,}400}-\frac{y^2}{25{,}600} = 1 \]
The ship lies somewhere on that hyperbola — on the branch nearer the station whose pulse arrived first. One measurement narrows an entire ocean to a single curve, which is a great deal but not a position.
A second pair of transmitters gives a second hyperbola, and the ship is where the two curves cross. This was the basis of the LORAN system used for marine navigation through most of the twentieth century, and the charts sailors used were printed with families of these hyperbolas drawn on them. Satellite navigation replaced it, but works on the same principle — differences of arrival times, intersected.
Commit first
Answer, then rate your confidence honestly.
Predict first
For a hyperbola, must a be larger than b?
Correct: No — a comes from the positive term, and b may well be larger.
\[ \tfrac{y^2}{4}-\tfrac{x^2}{25}=1: \; a = 2 < 5 = b \]
Why: Example 1 has a equal to 2 and b equal to 5, and its transverse axis is vertical because the y squared term is the positive one. For an ellipse the larger denominator marked the major axis, so a was necessarily the larger; for a hyperbola the SIGN does that job instead and the sizes are unconstrained. What is constrained is c, which always exceeds a because the relation adds rather than subtracts. Carrying the ellipse's rule across is the single most common error in this lesson.
Explain it
They have just met the ellipse and its two tacks and a string.
Discussion prompt
In four sentences or fewer, explain how a hyperbola differs from an ellipse.
Hint: Change one word in the string description.
Answer:
For an ellipse you mark every point whose two distances to the tacks ADD to a fixed total. For a hyperbola you mark every point whose two distances SUBTRACT to a fixed total.
That one change turns a closed loop into two separate curves that open away from each other and never close. They also run closer and closer to a pair of straight lines through the centre without ever touching them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For a, circle the positive term before reading anything. For c, remember that the foci sit outside the vertices, so c must be the largest. For asymptotes, draw the rectangle and read rise over run rather than recalling a formula. For mirrors, mark the centre on your sketch and notice that it is not on the curve.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a hyperbola page. Top left: draw a hyperbola with both foci marked, pick a point on each branch, and verify that both differences equal 2a. Top right: draw both orientations with their central rectangles, asymptotes and every part labelled, and write the two rules that differ from the ellipse — sign rather than size, and plus rather than minus. Middle: work Example 1 in full, drawing the rectangle first and the branches last, and mark the foci beyond the vertices. Bottom left: work Example 2 and both guided practice equations, noting beside each that c exceeded a. Bottom right: draw the mirror's cross section, mark the centre, the vertex and the rim, and write one sentence explaining why the height is a difference rather than a coordinate.
If any of your foci ended up between the vertices, recheck the relation: for a hyperbola c squared is a squared PLUS b squared, so c is always the largest of the three.
Recap
Five things, and the last of the four conic sections.
| If you see | Then |
|---|---|
| A minus between two squares equal to 1 | A hyperbola |
| The positive term | Its variable's axis is the transverse one, and its denominator is a squared |
| A rectangle 2a by 2b | Its diagonals are the asymptotes |
| Vertices at distance a | Foci at distance c, with c squared equal to a squared plus b squared |
| Foci and vertices given | b squared is c squared minus a squared |
| A physical object on one branch | Measure heights from the vertex, not the centre |
Lesson 9.6 translates all four conics away from the origin and shows how to classify any second-degree equation by its coefficients alone.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.5 Graph and Write Equations of Hyperbolas §9.5, pp. 642-647 — everything on these slides traces back here
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