9.05 Hyperbolas

Geometric definition of hyperbolas 

The hyperbola is one type of conic section, along with the circle, ellipse and parabola. We can construct a hyperbola given two points, $F_1$F1​ and $F_2$F2​, which we call the foci (singular focus).

The midpoint between the two foci is called the center. We label the distance from the center to each focus as $c$c. Our next step is to select some value $a$a, which will represent the distance from the center to each vertex $V_1$V1​ and $V_2$V2​ 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 $\left|PF_1-PF_2\right|=2a$|PF1​−PF2​|=2a.

$P$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 "$F_1$F1​" and "$F_2$F2​", but this doesn't mean that $F_1$F1​ has any more significance than $F_2$F2​.

Finally, let's draw a circle centered 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 triangle within the circle. This triangle has a hypotenuse of length $c$c and shorter sides $a$a and $b$b.

A right 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 $c^2=a^2+b^2$c2=a2+b2. 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.

Key characteristics

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

For any hyperbola, the two foci, the two vertices, and the center 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 $xy$xy-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: center ($C$C), two vertices ($V$V), two foci ($F$F), transverse and conjugate axes, asymptotes.

The behavior of the graph far from the center 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.

 

Standard form of hyperbolas

As we have seen with many different relations, we like to have a standard form which highlights transformations and allows us to quickly find key features to graph the relation.

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

Note:

• $a$a is the distance from the center to a vertex
• $b$b is the distance from a vertex to the closest asymptote
• $c$c is the distance from the center to a focus
• We have the relationship $c^2=a^2+b^2$c2=a2+b2 or $b^2=c^2-a^2$b2=c2−a2

The table below summarizes the characteristics of a translated hyperbola in both orientations:

Graph of $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1.   Orientation Horizontal Major Axis Center $\left(h,k\right)$(h,k) Foci $\left(h+c,k\right)$(h+c,k) and $\left(h-c,k\right)$(h−c,k) Vertices $\left(h+a,k\right)$(h+a,k) and $\left(k-a,k\right)$(k−a,k) Transverse axis $y=k$y=k Conjugate axis $x=h$x=h Asymptotes $y-k=\pm\frac{b}{a}(x-h)$y−k=±ba​(x−h)

Graph of $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1.   Orientation Vertical Major Axis Center $\left(h,k\right)$(h,k) Foci $\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,k−c) Vertices $\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,k−a) Transverse axis $x=h$x=h Conjugate axis $y=k$y=k Asymptotes $y-k=\pm\frac{a}{b}(x-h)$y−k=±ab​(x−h)

We can make a few more observations from this table. Firstly, consider a hyperbola centered 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)2a2​−(y−0)2b2​=1, which simplifies to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=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=√a2+b2, 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.

Here are some helpful diagrams with $a$a, $b$b and $c$c labeled.

Opening left and right:

The foci, vertices and asymptotes of a hyperbola.	The conjugate and transverse axes, with their lengths.

 

Opening up and down:

The foci, vertices and asymptotes of a hyperbola.	The conjugate and transverse axes, with their lengths.

 

Exploration

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)2a2​−(y−k)2b2​=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

Center: $\left(h,k\right)$(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=±ba​(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​

 

Worked example

Question 1

The hyperbola $\frac{x^2}{25}-\frac{y^2}{100}=1$x225​−y2100​=1 when translated $3$3 units to the right and $4$4 units up will be described by the equation $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x−3)225​−(y−4)2100​=1. Find the center, vertices, foci and asymptotes of the translated hyperbola. Then draw the graph of the hyperbola.

a) What is the center of the hyperbola?
Think: The equation of the hyperbola is in the form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1 and the center will be $\left(h,k\right)$(h,k).
Do: The center can just be read off from the equation as $\left(3,4\right)$(3,4).
Reflect: The values of $h$h and $k$k appear after a minus sign in the standard form of the equation. If the equation was in the form $\frac{\left(x+3\right)^2}{a^2}\ldots$(x+3)2a2​… then $h$h would equal $-3$−3.

 

b) What are the vertices of the hyperbola?
Think: For a hyperbola centered at $\left(h,k\right)$(h,k) and oriented horizontally, the vertices will be $a$a units horizontally in both directions from the center, $\left(h\pm a,k\right)$(h±a,k)
Do: From the equation $a^2=25$a2=25 so $a=5$a=5. The vertices will be at $\left(3\pm5,4\right)$(3±5,4), or written separately, $\left(-2,4\right),\left(8,4\right)$(−2,4),(8,4).
Reflect: Notice that the translated center, and the vertices all share the same $y$y-component. They will always lie on the transverse axis. 

Hyperbola centered at the origin and translated hyperbola with its vertices and center marked

c) What are the equations of the asymptotes of the hyperbola?
Think: For a hyperbola of the form above the asymptotes are in the form $y=\pm\frac{b}{a}\left(x-h\right)+k$y=±ba​(x−h)+k. We have found the values of $a$a, $h$h and $k$k. We need to find $b$b and substitute in the values.
Do: From the equation $b^2=100$b2=100 so $b=10$b=10. The asymptotes will be $y=\pm\frac{10}{5}\left(x-3\right)+4$y=±105​(x−3)+4, which simplifies to $y=\pm2\left(x-3\right)+4$y=±2(x−3)+4. 

d) Draw the graph of the hyperbola.
Think: The graph of the hyperbola is horizontally oriented, and will have a pair of asymptotes that intercept at the center $\left(3,4\right)$(3,4).
Do: The graph of the hyperbola is given below.
 

Graph of a hyperbola $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x−3)225​−(y−4)2100​=1

 

Practice questions

Question 2

Question 3

Question 4

 

Identify domain and range of hyperbolas

Recall that the domain of a relation is the set of $x$x-values in the relation. Graphically, we can think of the domain as all values of $x$x which correspond to one or more points in the relation.

Correspondingly, the range of a relation is the set of $y$y-values in the relation. Graphically, we can think of the range as all values of $y$y which correspond to one or more points in the relation.

It is often easiest to determine the domain and range of relations by looking at their graphs.

 

Hyperbolas with a horizontal transverse axis

A hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1 has a horizontal transverse axis and is centered at the origin.

We can see that the hyperbola has two branches; one on the right which corresponds to $x\ge a$x≥a, and one on the left which corresponds to $x\le-a$x≤−a. Values of $x$x between $-a$−a and $a$a do not correspond to any part of the hyperbola. So we have that:

Domain$=$=$\left(-\infty,-a\right]\cup\left[a,\infty\right)$(−∞,−a]∪[a,∞)

Looking at the other axis, we can see that each branch of the hyperbola covers all real values of $y$y. That is:

Range$=$=$\left(-\infty,\infty\right)$(−∞,∞)

 

Let's now look at a graph of $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1, which still has a horizontal transverse axis but is no longer centered at the origin.

This hyperbola is centered at the point $\left(h,k\right)$(h,k), with semi-major axis length $a$a. This means that the branch on the right corresponds to $x\ge h+a$x≥h+a, while the branch on the left corresponds to $x\le h-a$x≤h−a. Values of $x$x that are closer to the center than $a$a units (that is, values of $x$x between $h-a$h−a and $h+a$h+a) do not correspond to any part of the hyperbola. So we have that:

Domain$=$=$\left(-\infty,h-a\right]\cup\left[h+a,\infty\right)$(−∞,h−a]∪[h+a,∞)

Once again, the branches of the hyperbola still cover all real values of $y$y. So, as before, we have that:

Range$=$=$\left(-\infty,\infty\right)$(−∞,∞)

 

Hyperbolas with a vertical transverse axis

A hyperbola of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2​−x2b2​=1 has a vertical transverse axis instead.

The two branches of this hyperbola are separated vertically, where the top branch corresponds to $y\ge a$y≥a and the bottom one corresponds to $y\le-a$y≤−a. So for this orientation of hyperbola, it is the range that has two parts. On the other axis, we can see that all real values of $x$x correspond to both branches of the hyperbola. So we have that

Domain$=$=$\left(-\infty,\infty\right)$(−∞,∞),

and that

Range$=$=$\left(-\infty,-a\right]\cup\left[a,\infty\right)$(−∞,−a]∪[a,∞).

 

Here is a graph of $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1, which has been translated away from the origin.

This hyperbola is centered at the point $\left(h,k\right)$(h,k), with semi-major axis length $a$a. This means that the top branch corresponds to $y\ge k+a$y≥k+a, while the bottom branch corresponds to $y\le k-a$y≤k−a. Values of $y$y that are closer to the center than $a$a units (that is, values of $y$y between $k-a$k−a and $k+a$k+a) do not correspond to any part of the hyperbola. Meanwhile, all real values of $x$x still correspond to points on the hyperbola. So we have that

Domain$=$=$\left(-\infty,\infty\right)$(−∞,∞),

and that 

Range$=$=$\left(-\infty,k-a\right]\cup\left[k+a,\infty\right)$(−∞,k−a]∪[k+a,∞).

 

 

Practice questions

Question 5

Question 6

Find the equation of a hyperbola

We now examine the method of finding the equation of a hyperbola given certain identifying information.

Instead of using the values of $a$a, $b$b, $c$c, $h$h and $k$k to find the key features then graph, we are going to use the characteristics of the hyperbola graph to find $a$a, $b$b, $h$h and $k$k, then we simply need to fill them into the appropriate equation based on the direction of the transverse axis.

 

Worked examples

question 7

Find the equation of the hyperbola shown below.

 

Think: The center is at $\left(0,0\right)$(0,0) and it is aligned horizontally so it will have the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1. The vertices will be in the form $\left(\pm a,0\right)$(±a,0) and the foci will be at $\left(\pm c,0\right)$(±c,0) and we can find $b$b using the equation $c^2=a^2+b^2$c2=a2+b2.

Do: From the information above, and determining that a vertex is at $\left(5,0\right)$(5,0) and a focus is at $\left(13,0\right)$(13,0) we know that $a=5$a=5 and $c=13$c=13. We can then use the equation above:

$c^2$c2	$=$=	$a^2+b^2$a2+b2	(The equation relating $a$a, $b$b and $c$c)
$b^2$b2	$=$=	$c^2-a^2$c2−a2	(Rearranging the equation to make $b^2$b2 the subject)
$b$b	$=$=	$\sqrt{c^2-a^2}$√c2−a2	(Rearranging the equation to make $b$b the subject)
$b$b	$=$=	$\sqrt{13^2-5^2}$√132−52	(Substituting the values of $a$a and $b$b)
$b$b	$=$=	$\sqrt{144}$√144	(Evaluating the numbers inside the square root)
$b$b	$=$=	$12$12	(Evaluating the square root)

So the equation of the hyperbola will be $\frac{x^2}{5^2}-\frac{y^2}{12^2}=1$x252​−y2122​=1 or if we evaluate the denominators, $\frac{x^2}{25}-\frac{y^2}{144}=1$x225​−y2144​=1.

 

Question 8

Find the equation of the hyperbola that passes through the point $\left(12,5\right)$(12,5) and has a vertices at $\left(0,\pm3\right)$(0,±3).

Think: First to determine the orientation of the hyperbola we can look at the vertices. The vertices will lie on $y$y-axis as both $x$x-coordinates are the same. The vertices also tell us the value of $a$a. Using the equation and the point on the hyperbola we can determine the value of $b$b.

 

Do: The value of $a$a will be $3$3. The equation of the hyperbola will be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2​−x2b2​=1. If we substitute the value of $a$a and the point's $x$x and $y$y-coordinates for $x$x and $y$y in the equation.

$\frac{y^2}{a^2}-\frac{x^2}{b^2}$y2a2​−x2b2​	$=$=	$1$1	(Equation of hyperbola)
$\frac{x^2}{b^2}$x2b2​	$=$=	$\frac{y^2}{a^2}-1$y2a2​−1	(Making the term with $b$b the subject)
$\frac{x^2}{b^2}$x2b2​	$=$=	$\frac{y^2-a^2}{a^2}$y2−a2a2​	(Combining two terms)
$b^2$b2	$=$=	$\frac{x^2a^2}{y^2-a^2}$x2a2y2−a2​	(Making $b^2$b2 the subject)
$b$b	$=$=	$\frac{xa}{\sqrt{y^2-a^2}}$xa√y2−a2​	(Taking the square root of both sides)
$b$b	$=$=	$\frac{12\times3}{\sqrt{5^2-3^2}}$12×3√52−32​	(Substituting)
$b$b	$=$=	$\frac{36}{4}$364​	(Simplifying the numerator and denominator)
$b$b	$=$=	$9$9	(Simplifying the fraction)

So the equation of the hyperbola will be $\frac{y^2}{3^2}-\frac{x^2}{9^2}=1$y232​−x292​=1 or if we evaluate the denominators $\frac{y^2}{9}-\frac{x^2}{81}=1$y29​−x281​=1.

 

Question 9

Find the equation of the hyperbola with the asymptotes $y=\pm2\left(x-3\right)+4$y=±2(x−3)+4, and vertices at $\left(-2,4\right)$(−2,4) and $\left(8,4\right)$(8,4).

Think: From the asymptotes and vertices we can tell that this hyperbola is not centered at the origin. From the vertices we can also tell that the hyperbola is oriented horizontally so its asymptotes will be in the form $y=\pm\frac{b}{a}\left(x-h\right)+k$y=±ba​(x−h)+k. From this we can read off $h$h and $k$k. The distance between the two vertices will be $2a$2a. We can use the slope of the asymptote to determine $b$b.

 

Do: The value of $h=3$h=3 and $k=4$k=4. The distance between the two vertices will be $2a=\left|8-\left(-2\right)\right|$2a=|8−(−2)| so $a=5$a=5.

Now the slope $\frac{b}{a}$ba​ should be equal to $2$2. So we can use this and the value of $a$a to find $b$b.

$\frac{b}{a}$ba​	$=$=	$2$2
$\frac{b}{5}$b5​	$=$=	$2$2
$b$b	$=$=	$2\times5$2×5
$b$b	$=$=	$10$10

So the equation of the hyperbola will be $\frac{\left(x-3\right)^2}{5^2}-\frac{\left(y-4\right)^2}{10^2}=1$(x−3)252​−(y−4)2102​=1 or if we evaluate the denominators, $\frac{\left(x-3\right)^2}{25}-\frac{\left(y-4\right)^2}{100}=1$(x−3)225​−(y−4)2100​=1.

Reflect: Notice that we also could have found the values of $h$h and $k$k using the vertices. The coordinate in the middle of the two points will be the center. So in this case the point between the vertices will be $\left(\frac{-2+8}{2},4\right)$(−2+82​,4) which simplifies to $\left(3,4\right)$(3,4). These are values we found for $h$h and $k$k.

 

Practice questions

Question 10

Question 11

Question 12

 

Manipulate an equation to arrive at standard form 

Typically we are given the equation of a hyperbola in the standard form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1 or $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1. Ideally, we would like to express any equation of a hyperbola in this form since we can immediately identify its center, orientation, and other features.

But generally speaking, the equation of a hyperbola can be expressed in many ways. For instance, the following pair of equations both represent the same hyperbola:

$4\left(x-2\right)^2-25\left(y-3\right)^2=100$4(x−2)2−25(y−3)2=100 and $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105

In order to express the above equations in the standard form, we can manipulate both sides by multiplying or by completing the square.

 

Worked example

question 13

Consider the equation of the hyperbola $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105.

Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1.

Think: Completing the square allows us to rewrite the quadratic expressions in $x$x and $y$y to be perfect squares of the form $\left(x-h\right)^2$(x−h)2 and $\left(y-k\right)^2$(y−k)2. Before we do so, it will be more convenient if the coefficient of $y^2$y2 is $1$1.

$x^2-4x-\left(y^2-6y\right)=105$x2−4x−(y2−6y)=105

Then to complete the square in $x$x we look at the expression $x^2-4x$x2−4x, halve and square the coefficient of $x$x, and then add the result to both sides of the equation.

Do: Halving and squaring $-4$−4 gives a result of $4$4, so we add this result to both sides of the equation:

$x^2-4x+4-\left(y^2-6y\right)=105+4$x2−4x+4−(y2−6y)=105+4

Similarly, to complete the square in $y$y we add $9$9 inside the parentheses:

$x^2-4x+4-\left(y^2-6y+9\right)=105+4-9$x2−4x+4−(y2−6y+9)=105+4−9

Because of the negative sign outside of the parentheses, we will subtract $9$9 from the other side to balance the equation.

Putting this together, we get the following steps:

$x^2-4x-y^2+6y$x2−4x−y2+6y	$=$=	$105$105	(Writing down the equation)
$x^2-4x-\left(y^2-6y\right)$x2−4x−(y2−6y)	$=$=	$105$105	(Factoring the terms with $y$y)
$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9)	$=$=	$105+4-9$105+4−9	(Completing the square for $x$x and $y$y)
$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9)	$=$=	$100$100	(Simplifying the constant terms)
$\left(x-2\right)^2-\left(y-3\right)^2$(x−2)2−(y−3)2	$=$=	$100$100	(Factoring the perfect squares)
$\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}$(x−2)2100​−(y−3)2100​	$=$=	$1$1	(Dividing both sides by $100$100)

The equation of the hyperbola is then $\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}=1$(x−2)2100​−(y−3)2100​=1 which is in the form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1.

Reflect: The hyperbola with equation $x^2-4x-y^2+6y=105$x2−4x−y2+6y=105 has a center of $\left(2,3\right)$(2,3) with vertices $\left(-8,3\right)$(−8,3) and $\left(12,3\right)$(12,3). What are the coordinates of the foci?

 

question 14

Consider the equation of the circle $3y^2+12y-x^2-6x=3$3y2+12y−x2−6x=3.

Rewrite the equation in the standard form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1.

Think: Completing the square allows us to rewrite the terms containing $x$x and $y$y in the form $\left(x-h\right)^2$(x−h)2 and $\left(y-k\right)^2$(y−k)2. Before we do so, it will be more convenient if the coefficients of $x^2$x2 and $y^2$y2 were $1$1.

Do: In this case, the coefficient of $x^2$x2 is $-1$−1 and the coefficient of $y^2$y2 is $3$3, so we factor out these coefficients:

$3\left(y^2+4y\right)-\left(x^2+6x\right)=3$3(y2+4y)−(x2+6x)=3

Now we can complete the square in $x$x and $y$y:

$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)=3+3\times4-9$3(y2+4y+4)−(x2+6x+9)=3+3×4−9

Notice that we've added by $3\times4$3×4 and subtracted by $9$9 on the other side to balance the equation. This is needed because the terms that we've added on the left hand side are being multiplied by $3$3 and $-1$−1 respectively.

Putting this together, we get the following steps:

$3y^2+12y-x^2-6x$3y2+12y−x2−6x	$=$=	$3$3	(Writing down the equation)
$3\left(y^2+4y\right)-\left(x^2+6x\right)$3(y2+4y)−(x2+6x)	$=$=	$3$3	(Factoring the terms with $x$x and $y$y)
$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9)	$=$=	$3+3\times4-9$3+3×4−9	(Completing the square for $x$x and $y$y)
$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9)	$=$=	$6$6	(Simplifying the constant terms)
$3\left(y+2\right)^2-\left(x+3\right)^2$3(y+2)2−(x+3)2	$=$=	$6$6	(Factoring the perfect squares)
$\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}$(y+2)22​−(x+3)26​	$=$=	$1$1	(Dividing both sides by $6$6)

So the equation of the hyperbola can be expressed in the form $\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}=1$(y+2)22​−(x+3)26​=1 which is in the form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(h-k\right)^2}{b^2}=1$(y−k)2a2​−(h−k)2b2​=1.

Reflect: The center of the hyperbola is $3y^2+12y-x^2-6x=3$3y2+12y−x2−6x=3 is $\left(-3,-2\right)$(−3,−2) and has vertices $\left(-3,-2+\sqrt{2}\right)$(−3,−2+√2) and $\left(-3,-2-\sqrt{2}\right)$(−3,−2−√2). What are the coordinates of the foci?

 

Practice questions

question 15

question 16

question 17

 

Applications of hyperbolic functions

We have now seen how to graph and identify the graphs of $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1 and $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(h-k\right)^2}{b^2}=1$(y−k)2a2​−(h−k)2b2​=1. We now want to look at some applications of these relations.

Worked example

Question 18

Two stationary aircraft carriers sit somewhere in the Pacific Ocean. The carrier HMS Mawson lies $10$10 nautical miles due east of the carrier HMS Scott. They each receive a distress signal from a jet plane and fear that it has tragically crashed into the sea.

Using the time delay between the ships receiving the distress signals, navigators determined that the plane must have been $6$6 nautical miles closer to HMS Scott than HMS Mawson when it sent the signal.

The navigators are tasked with defining a search zone, and preparations are immediately made to send out a search plane.

 

a) What are the locations where the difference in the distances to each carrier is $6$6 nautical miles?

Think: Suppose that $d_1$d1​ and $d_2$d2​ are the two distances between a point $P\left(x,y\right)$P(x,y) and the ships. What is the locus of all points $P$P that satisfy the condition $d_1-d_2=6$d1​−d2​=6 nautical miles (irrespective of any other condition on $P$P)?

This locus defines a hyperbola. You can check this for yourself by using the distance equation $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$d=√(x2​−x1​)2+(y2​−y1​)2 and rearranging into the equation of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1. From this the navigators placed the ships on a coordinate system and determined the possible locations for the plane seen in the applet below.

Notice that, in the top box, for all points along the hyperbola the difference in distance to the two ships is $6$6 nautical miles.

Do: We would like to find the equation that describes the hyperbola above in the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1. To do this we will find the values of $a$a and $b$b.

The value of $a$a is the half of the length of the transverse axis. For any point on the hyperbola the difference in distances to each foci from that point is $2a$2a. That is, looking at the applet above we can see that $\left|d_1-d_2\right|$|d1​−d2​| is equal to $2a$2a. Since $\left|d_1-d_2\right|$|d1​−d2​| is defined to be $6$6 for this hyperbola, we have that $a=\frac{6}{2}=3$a=62​=3.

The value of $b$b is half the length of the conjugate axis, which we don't know. We do know the focal length $c$c, however, which is linked to $a$a and $b$b by the equation $c^2=a^2+b^2$c2=a2+b2. Rearranging to make $b$b the subject, we have $b=\sqrt{c^2-a^2}=\sqrt{5^2-3^2}=\sqrt{16}=4$b=√c2−a2=√52−32=√16=4.

So we now know that the equation of this hyperbola is given by $\frac{x^2}{3^2}-\frac{y^2}{4^2}=1$x232​−y242​=1.

 

b) The navigators determine that the plane was heading directly for HMS Mawson with a bearing of $045^\circ$045°. What is the most likely location of the plane?

Think: A straight line with a bearing of $045^\circ$045° has a slope of $1$1. Using our coordinate system, this means that the plane was traveling along the line $y=x-5$y=x−5 (which has a slope of $1$1 and passes through the location of HMS Mawson).

Do: By solving the two equations simultaneously, we can pinpoint the most likely location of the plane:

$\frac{x^2}{9}-\frac{\left(x-5\right)^2}{16}$x29​−(x−5)216​	$=$=	$1$1	by substituting $y=x-5$y=x−5 into the equation of the hyperbola.
$16x^2-9\left(x-5\right)^2$16x2−9(x−5)2	$=$=	$144$144
$16x^2-9x^2+90x-225$16x2−9x2+90x−225	$=$=	$144$144
$7x^2+90x-369$7x2+90x−369	$=$=	$0$0
$\therefore x$∴x	$=$=	$\frac{-90\pm\sqrt{8100+10332}}{14}$−90±√8100+1033214​
	$\approx$≈	$-16.126,3.269$−16.126,3.269

 

Reflect: We can reject the value $x=3.269$x=3.269 as a solution because we know that $d_1>d_2$d1​>d2​ (that is, we know that the plane was closer to HMS Scott, which received the signal first). Through the process of developing the hyperbola's equation, the squaring of quantities allowed the case where $d_2-d_1=6$d2​−d1​=6, but we only want to consider the left branch of the hyperbola (the branch that is closer to HMS Scott).

As such, the most likely location of the plane (according to the coordinate system) is $\left(-16.126,-21.126\right)$(−16.126,−21.126). This is shown in the diagram below, along with the rejected solution shown in blue.

The distance from HMS Mawson is $d_1=29.88$d1​=29.88 nautical miles, in the direction $S45^\circ W$S45°W from HMS Mawson.

Practice question

Question 19