Hong Kong

Stage 4 - Stage 5

Lesson

The hyperbola is one type of conic section, along with the circle, ellipse and parabola. We can construct a hyperbola given two points, $F_1$`F`1 and $F_2$`F`2, which we call the foci (singular focus).

The midpoint between the two foci is called the centre. We label the distance from the centre to each focus as $c$`c`. Our next step is to select some value $a$`a`, which will represent the distance from the centre to each vertex $V_1$`V`1 and $V_2$`V`2 of the hyperbola.

Once we have the two foci and the value for $a$`a`, we define the hyperbola to be the curve consisting of all the points $P$`P` that satisfy the relationship $\left|PF_1-PF_2\right|=2a$|`P``F`1−`P``F`2|=2`a`.

$P$P is a point that satisfies $\left|PF_1-PF_2\right|=2a$|PF1−PF2|=2a. |

The absolute value in the definition reflects the fact that both foci play an identical role in the construction of the curve. We use the labels "$F_1$`F`1" and "$F_2$`F`2", but this doesn't mean that $F_1$`F`1 has any more significance than $F_2$`F`2.

Finally, let's draw a circle centred at $C$`C`, with radius $c$`c`. This circle will pass through both foci. If we draw a tangent to the hyperbola at a vertex, we will have constructed a right-angled triangle within the circle. This triangle has a hypotenuse of length $c$`c` and shorter sides $a$`a` and $b$`b`.

A right-angled triangle with hypotenuse $c$c and shorter sides $a$a and $b$b. |

If we label the remaining side $b$`b`, then we arrive at the relationship $c^2=a^2+b^2$`c`2=`a`2+`b`2. This will be useful when we try to understand the properties of a hyperbola using only its equation, which will be given in terms of $a$`a` and $b$`b`.

Now we can explore some of the characteristics of hyperbolas.

The midpoint between the two foci is called the centre. The hyperbola also has two vertices, which are the points on the graph that are closest to the centre.

For any hyperbola, the two foci, the two vertices, and the centre will all lie on the same straight line, called the transverse axis. At this stage we will concentrate on hyperbolas whose transverse axis is a horizontal line or a vertical line in the $xy$`x``y`-plane.

A hyperbola with a horizontal transverse axis can be said to open to the left and right, while a hyperbola with a vertical transverse axis opens in the up and down direction.

The key components of a hyperbola: centre ($C$C), two vertices ($V$V), two foci ($F$F),transverse and conjugate axes, asymptotes. |

The behaviour of the graph far from the centre is described by the two asymptotes, which are the lines that the graph approaches but never reaches. In general the two asymptotes are not perpendicular.

All of this information is captured in the equation for the hyperbola, and is summarised below for the two main orientations we will be considering.

Characteristics of hyperbolas

Centre: $\left(h,k\right)$( Vertices: $\left(h\pm a,k\right)$( $c^2=a^2+b^2$ Foci: $\left(h\pm c,k\right)$( Transverse axis: $y=k$ Conjugate axis: $x=h$ Asymptotes: $y-k=\pm\frac{b}{a}\left(x-h\right)$ |
Centre: $\left(h,k\right)$( Vertices: $\left(h,k\pm a\right)$( $c^2=a^2+b^2$ Foci: $\left(h,k\pm c\right)$( Transverse axis: $x=h$ Conjugate axis: $y=k$ Asymptotes: $y-k=\pm\frac{a}{b}\left(x-h\right)$ |

We can make a few more observations from this table. Firstly, consider a hyperbola centred at the origin with a horizontal transverse axis. It has the equation $\frac{\left(x-0\right)^2}{a^2}-\frac{\left(y-0\right)^2}{b^2}=1$(`x`−0)2`a`2−(`y`−0)2`b`2=1, which simplifies to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=1. The expressions for the other key characteristics will simplify in a similar way.

Secondly, the transverse axis is perpendicular to the conjugate axis. Both of these lines act as axes of symmetry of the graph.

Finally, given the value of $a$`a` and $b$`b` from the equation we can define a third variable $c=\sqrt{a^2+b^2}$`c`=√`a`2+`b`2, and we can see that this plays an important role in the coordinates of the foci. We can see that the value of $h$`h` and $k$`k` influence the **location** of the graph, while the value of $a$`a` and $b$`b` (and so $c$`c`) influence the **shape** of the graph.

Let's determine the key characteristics of the hyperbola given by the equation $\frac{\left(x+1\right)^2}{16}-\frac{\left(y-6\right)^2}{9}=1$(`x`+1)216−(`y`−6)29=1.

Looking at the equation, the term involving $y$`y` is being subtracted from the term involving $x$`x`, which means this hyperbola has a horizontal transverse axis and opens to the left and right. Here is the graph of the hyperbola.

Graph of $\frac{\left(x+1\right)^2}{16}-\frac{\left(y-6\right)^2}{9}=1$(x+1)216−(y−6)29=1. |

The equation has the form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(`x`−`h`)2`a`2−(`y`−`k`)2`b`2=1, where:

- $h=-1$
`h`=−1 - $k=6$
`k`=6 - $a=4$
`a`=4 - $b=3$
`b`=3

With $c=\sqrt{4^2+3^2}=5$`c`=√42+32=5, we can use these five values to find the key characteristics of the hyperbola as follows.

Orientation: Horizontal transverse axis

Centre: $\left(h,k\right)$(`h`,`k`) $\rightarrow$→ $\left(-1,6\right)$(−1,6)

Vertices: $\left(h\pm a,k\right)$(`h`±`a`,`k`) $\rightarrow$→ $\left(-1\pm4,6\right)$(−1±4,6), which gives $\left(3,6\right)$(3,6) and $\left(-5,6\right)$(−5,6)

Foci: $\left(h\pm c,k\right)$(`h`±`c`,`k`) $\rightarrow$→ $\left(-1\pm5,6\right)$(−1±5,6), which gives $\left(4,6\right)$(4,6) and $\left(-6,6\right)$(−6,6)

Transverse axis: $y=k$`y`=`k` $\rightarrow$→ $y=6$`y`=6

Conjugate axis: $x=h$`x`=`h` $\rightarrow$→ $x=-1$`x`=−1

Asymptotes: $y-k=\pm\frac{b}{a}\left(x-h\right)$`y`−`k`=±`b``a`(`x`−`h`) $\rightarrow$→ $y-6=\pm\frac{3}{4}\left(x+1\right)$`y`−6=±34(`x`+1), which gives $y=\frac{3}{4}x+\frac{27}{4}$`y`=34`x`+274 and $y=-\frac{3}{4}x+\frac{21}{4}$`y`=−34`x`+214

The graph of $\frac{x^2}{16}-\frac{y^2}{9}=1$`x`216−`y`29=1 is shown below.

Loading Graph...

What are the coordinates of the centre of the hyperbola?

What are the coordinates of the vertices? Write each pair of coordinates on the same line, separated by a comma.

What are the coordinates of the foci? Write each pair of coordinates on the same line, separated by a comma.

What is the orientation of the graph?

Horizontal transverse axis

AVertical transverse axis

BWhat is the equation of the transverse axis?

What is the equation of the conjugate axis?

What are the equations of the asymptotes? Write each equation on the same line, separated by a comma.

Consider the hyperbola with the equation $\frac{y^2}{9}-\frac{x^2}{25}=1$`y`29−`x`225=1.

Based on the equation, what is the orientation of the graph of this hyperbola?

Horizontal transverse axis

AVertical transverse axis

BWhat are the coordinates of the vertices? Write each pair of coordinates on the same line, separated by a comma.

What are the coordinates of the foci? Write each pair of coordinates on the same line, separated by a comma.

What are the equations of the asymptotes? Write each equation on the same line, separated by a comma.

Consider the hyperbola with the equation $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-2\right)^2}{9}=1$(`x`−1)24−(`y`−2)29=1.

Based on the equation, what is the orientation of the graph of this hyperbola?

Horizontal transverse axis

AVertical transverse axis

BWhat are the coordinates of the centre?

Find the distance $c$

`c`between the centre and a focus of the hyperbola.What are the coordinates of the foci? Write each pair of coordinates on the same line, separated by a comma.

What is the equation of the transverse axis?

What is the equation of the conjugate axis?

What are the equations of the asymptotes? Write each equation in the form $y=mx+c$

`y`=`m``x`+`c`, separated by a comma.