A hyperbola is the curve made of all points whose distances to two foci differ by a fixed amount; it splits into two branches.
An ellipse keeps the sum of the distances fixed and a hyperbola keeps the difference fixed, which makes the pair easy to remember.
The two branches get closer and closer to two lines, the asymptotes, but never meet them.
Move a, b and the point P: PF′ − PF always comes out as 2a, and the diagonals of the 2a by 2b rectangle line up with the asymptotes.
Definition of a Hyperbola
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The foci, vertices and asymptotes of x²/a² − y²/b² = 1, with a point P on the right branch
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👀 What to read in the picture
①The two red points are the foci F′(−c, 0) and F(c, 0), with c² = a² + b²
②The top band shows the distances from P on the right branch to the two foci, and their difference PF′ − PF is 2a wherever P is
③For a point on the left branch, PF − PF′ = 2a, so for both branches together |PF′ − PF| = 2a
④The gold points are the vertices (±a, 0), and the dashed lines are the asymptotes y = ±(b/a)x
Asymptotes and the Rectangle
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The 2a by 2b rectangle and the asymptotes along its extended diagonals
📐 How to draw the asymptotes
①Draw a rectangle 2a wide and 2b tall, centered at the origin
②Extend its two diagonals forever: they are the asymptotes y = ±(b/a)x
③The hyperbola gets arbitrarily close to the asymptotes but never meets them
④The vertices (±a, 0) lie on the left and right sides of the rectangle
Standard Form
Standard form (transverse axis along the x-axis)
x²a² − y²b² = 1
Foci (±c, 0), asymptotes y = ±(b/a)x
Standard form (transverse axis along the y-axis)
y²a² − x²b² = 1
Foci (0, ±c), asymptotes y = ±(a/b)x
🔍 Compared with the ellipse
①An ellipse adds the two terms, so it is a closed curve
②A hyperbola subtracts them, so it is open, with two branches
③For an ellipse, c² = a² − b²
④For a hyperbola, c² = a² + b², whichever of a and b is larger
Foci and Eccentricity
Focal relationship
c² = a² + b²
(center-to-focus distance)² = a² + b²
Eccentricity
e = ca > 1
A small b/a gives e close to 1 and narrow branches; a large b/a gives a larger e and branches that spread wide
🎯 What eccentricity means
①The eccentricity of a hyperbola is always e > 1
②e = √(1 + b²/a²)
③The larger b/a is, the steeper the asymptotes and the larger e
④A rectangular hyperbola (a = b) has e = √2 and asymptotes y = ±x
Worked Examples
Example 1
Find the coordinates of the foci and the equations of the asymptotes of the hyperbola x²/9 − y²/16 = 1.
1
From a² = 9 and b² = 16, a = 3 and b = 4; find the foci from c² = a² + b².
c² = 9 + 16 = 25, c = 5
2
The transverse axis is the x-axis, so the foci lie on the x-axis and the asymptotes are y = ±(b/a)x.
foci (±5, 0), asymptotes y = ±43x
▸ Foci (±5, 0), asymptotes y = ±(4/3)x
For a hyperbola c² = a² + b² adds the squares, the opposite sign from c² = a² − b² for an ellipse.
Example 2
Find the equation of the hyperbola with foci (±5, 0) whose distances to the foci differ by 6.
1
From the difference 2a = 6, a = 3, and from the foci, c = 5.
2a = 6 ⇒ a = 3, c = 5
2
Find b² from b² = c² − a² and substitute into the standard form.
b² = 25 − 9 = 16, x²9 − y²16 = 1
▸ x²/9 − y²/16 = 1
Fixing a first from the definition |PF′ − PF| = 2a makes b² follow from c² − a².
Summary
Key comparison
x²a² − y²b² = 1, c² = a² + b²
The minus sign marks a hyperbola, and c is always larger than a
2021 KICE mock exam Math (Geometry) type, adapted
What is the distance between the two foci of the hyperbola x²/4 − y²/5 = 1?
①6
②5
③4
④3
⑤2
▸ ① 6
1
Since a² = 4 and b² = 5, c² = a² + b² = 4 + 5 = 9, so c = 3.
2
The distance between the foci is 2c.
2c = 2 × 3 = 6
🎯 Exam Points
①The definition is |PF′ − PF| = 2a
②c² = a² + b², the opposite sign from the ellipse
③The asymptotes y = ±(b/a)x come up often
④The eccentricity satisfies e > 1
⑤By eccentricity: ellipse 0 < e < 1, parabola e = 1, hyperbola e > 1