Hyperbola

A hyperbola is a curve split into two branches, made of all the points where the difference of the distances to two foci stays constant. If an ellipse is the 'sum of distances is constant' curve, then a hyperbola is its partner: the 'difference of distances is constant' curve. Its two branches get closer and closer to straight lines called asymptotes but never actually touch them. Drag the a and b sliders and check that the difference of the distances to the two foci stays fixed at 2a, and that the diagonals of the basic rectangle become exactly the asymptotes.

Definition of a Hyperbola

a (semi-transverse axis)2

b (semi-conjugate axis)1.5

👀 What is a Hyperbola?

①Two foci F, F' exist

②For any point P on the hyperbola, |PF' − PF| = 2a (constant absolute difference)

③Ellipse: 'sum is constant', Hyperbola: 'difference is constant'

④The curve has two branches

Asymptotes and the Rectangle

📐 Secret of Asymptotes

①Draw a rectangle of width 2a, height 2b centered at origin

②Extending its diagonals infinitely gives the asymptotes

③The hyperbola gets arbitrarily close to but never touches the asymptotes

④Asymptotes: y = ±(b/a)x

Standard Form

Standard Form (Horizontal)

x²a² − y²b² = 1

Foci (±c, 0), asymptotes y = ±(b/a)x

Standard Form (Vertical)

y²a² − x²b² = 1

Foci (0, ±c), asymptotes y = ±(a/b)x

🔍 Comparison with Ellipse

①Ellipse: + sign → closed curve

②Hyperbola: − sign → open curve (two branches)

③Ellipse: c² = a² − b² (a > b)

④Hyperbola: c² = a² + b² (no order between a, b)

Foci and Eccentricity

Focus Relation

c² = a² + b²

focal distance² = transverse² + conjugate²

Eccentricity

e = ca > 1

e→1: narrow asymptotes; large e: wide asymptotes

🎯 Eccentricity Intuition

①Hyperbola eccentricity is always e > 1

②e = √(1 + b²/a²)

③Larger b/a → steeper asymptotes → larger e

④Rectangular hyperbola (a=b): e = √2, asymptotes y = ±x

Work It Out

Example 1

Find the coordinates of the foci and the equations of the asymptotes of the hyperbola x²/9 − y²/16 = 1.

From a²=9, b²=16 we get a=3, b=4. Find the foci using c²=a²+b².

c² = 9 + 16 = 25, c = 5

It is horizontal, 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 (addition). Note the sign is opposite to the ellipse c² = a² − b².

Example 2

Find the equation of the hyperbola whose foci are (±5, 0) and whose difference of distances to the two foci is 6.

The difference 2a = 6 gives a = 3, and from the foci c = 5.

2a = 6 ⇒ a = 3, c = 5

Find b² from b² = c² − a² and substitute into the standard form.

b² = 25 − 9 = 16, x²9 − y²16 = 1

▸ x²/9 − y²/16 = 1

Starting from the definition |PF′ − PF| = 2a fixes a first, then b² follows as c² − a².

Summary

Key Comparison

x²a² − y²b² = 1, c² = a² + b²

'−' sign is the hallmark of a hyperbola | c is always greater 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

Since a²=4, b²=5, we have c² = a² + b² = 4 + 5 = 9, so c = 3.

The distance between the two foci is 2c.

2c = 2 × 3 = 6

🎯 Exam Points

①Definition: |PF' − PF| = 2a

②c² = a² + b² (sign opposite to ellipse!)

③Asymptotes y = ±(b/a)x — frequently tested

④Eccentricity e > 1

⑤Comparing eccentricities: ellipse(0<e<1), parabola(e=1), hyperbola(e>1)

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