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Stage 4 - Stage 5

Lesson

A horizontal hyperbola is defined by the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=1 in standard form. Some of its key properties are:

- The vertices $\left(\pm a,0\right)$(±
`a`,0), which are the turning points of the two branches of the hyperbola. - The foci $\left(\pm c,0\right)$(±
`c`,0), where $c=\sqrt{a^2+b^2}$`c`=√`a`2+`b`2. - The transverse axis, which connects the vertices. It has length $2a$2
`a`. - The conjugate axis, which is perpendicular to the conjugate axis with length $2b$2
`b`. - The asymptotes, which are the two lines that bound the hyperbola. They are defined by $y=\pm\frac{b}{a}x$
`y`=±`b``a``x`.

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

A vertical hyperbola is defined by the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$`y`2`a`2−`x`2`b`2=1 in standard form. This looks like the equation of a horizontal hyperbola with the $x$`x`- and $y$`y`-values switched. As such, the foci, vertices and asymptotes can be found by interchanging the $x$`x`- and $y$`y`-values, as seen in the diagrams below.

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

A horizontal hyperbola translated $h$`h` units horizontally and $k$`k` units vertically has the equation $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(`x`−`h`)2`a`2−(`y`−`k`)2`b`2=1.

Its centre will be at the point $\left(h,k\right)$(`h`,`k`), and its asymptotes are the lines $y=\pm\frac{b}{a}\left(x-h\right)+k$`y`=±`b``a`(`x`−`h`)+`k`.

Centre of a hyperbola |

A vertical hyperbola can be translated similarly, and will have an equation of the form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(`y`−`k`)2`a`2−(`x`−`h`)2`b`2=1.

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 $d_1$`d`1 and $d_2$`d`2 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 $d_1-d_2=6$`d`1−`d`2=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`=√(`x`2−`x`1)2+(`y`2−`y`1)2 and rearranging into the equation of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=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$2`a`. That is, looking at the applet above we can see that $\left|d_1-d_2\right|$|`d`1−`d`2| is equal to $2a$2`a`. Since $\left|d_1-d_2\right|$|`d`1−`d`2| 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 $c^2=a^2+b^2$`c`2=`a`2+`b`2. Rearranging to make $b$`b` the subject, we have $b=\sqrt{c^2-a^2}=\sqrt{5^2-3^2}=\sqrt{16}=4$`b`=√`c`2−`a`2=√52−32=√16=4.

So we now know that the equation of this hyperbola is given by $\frac{x^2}{3^2}-\frac{y^2}{4^2}=1$`x`232−`y`242=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 gradient of $1$1. Using our coordinate system, this means that the plane was travelling along the line $y=x-5$`y`=`x`−5 (which has a gradient 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 $d_1>d_2$`d`1>`d`2 (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 $d_2-d_1=6$`d`2−`d`1=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 $d_1=29.88$`d`1=29.88 nautical miles, in the direction $S45^\circ W$`S`45°`W` from HMS Mawson.

The technique of locating an object based on time delays is well established. In fact if there are three stations receiving the distress signal, two hyperbolas can be generated from any two of the three possible pairings of time differences available. The object will then be located at the intersection of the two hyperbolas.

Using the applet below move the ships and plane to investigate how using three ships, we can produce two hyperbolas that intersect revealing the planes location. Notice that four possible points arise, but we can use a similar technique as above to choose the point based on which signal is received first.

An astronomer is studying the remains of an old star that has ejected its outer atmosphere in bursts of material. A cross section of the nebula has the shape of a hyperbola as plotted below. The units are given in light years. The point $Q$`Q`$\left(0.6,0.2\right)$(0.6,0.2) is on the asymptote of the gas shells. The point $V$`V`$\left(1.2,0\right)$(1.2,0) is the vertex of one of the gas shells.

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One asymptote passes through the point $Q$

`Q`$\left(0.6,0.2\right)$(0.6,0.2) as shown in the diagram. Find the equation of this asymptote.What is the length of the transverse axis?

What is the length of the conjugate axis?

Find the equation of the hyperbola describing the shape of the nebula in standard form.

Find the $x$

`x`-values of the foci. Each value should be rounded to two decimal places.