New Zealand
Level 8 - NCEA Level 3

# Applications of Ellipses

Lesson

Below are two images with the important features of an ellipse:

 Horizontally aligned ellipse Vertically aligned ellipse

The standard form for a central ellipse depends on the orientation of the ellipse.  The equations and attributes are summarized in the table below, given the following:

• $a$a is the length of the semi-major axis.
• $b$b is the length of the semi-minor axis.
• $c$c is the distance from the center to each focus.

Notice that by this definition, it is always true that $a>b$a>b. It is also true that the parameters $a$a$b$b, and $c$c have the relationship $c^2=a^2-b^2$c2=a2b2.

Orientation Horizontal Major Axis Vertical Major Axis
Standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2+y2b2=1 $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2+y2a2=1
Center $\left(0,0\right)$(0,0) $\left(0,0\right)$(0,0)
Foci $\left(c,0\right)$(c,0) and $\left(-c,0\right)$(c,0) $\left(0,c\right)$(0,c) and $\left(0,-c\right)$(0,c)
Vertices $\left(a,0\right)$(a,0) and $\left(-a,0\right)$(a,0) $\left(0,a\right)$(0,a) and $\left(0,-a\right)$(0,a)
Co-vertices $\left(0,b\right)$(0,b) and $\left(0,-b\right)$(0,b) $\left(b,0\right)$(b,0) and $\left(-b,0\right)$(b,0)
Major axis $y=0$y=0 $x=0$x=0
Minor axis $x=0$x=0 $y=0$y=0

### Translated ellipses

If an ellipse is translated horizontally or vertically from the origin, the parameter $a$a, $b$b, and $c$c still have the same meaning. However, we must take into account that the centre of the ellipse has moved.  Given the following definitions for $h$h and $k$k,

• $h$h denotes the translation in the horizontal direction from $\left(0,0\right)$(0,0).
• $k$k denotes the translation in the vertical direction from $\left(0,0\right)$(0,0).
 An ellipse translated horizontally by $h$h and vertically by $k$k.

The table below summarises the standard form of an ellipse in both orientations.

Orientation Horizontal Major Axis Vertical Major Axis
Standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)2a2=1
Center $\left(h,k\right)$(h,k) $\left(h,k\right)$(h,k)
Foci $\left(h+c,k\right)$(h+c,k) and $\left(h-c,k\right)$(hc,k) $\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,kc)
Vertices $\left(h+a,k\right)$(h+a,k) and $\left(h-a,k\right)$(ha,k) $\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,ka)
Co-vertices $\left(h,k+b\right)$(h,k+b) and $\left(h,k-b\right)$(h,kb) $\left(h+b,k\right)$(h+b,k) and $\left(h-b,k\right)$(hb,k)
Major axis $y=k$y=k $x=h$x=h
Minor axis $x=h$x=h $y=k$y=k

Essentially, the information is the same as the central ellipse.  But the values of $h$h and $k$k are added to the $x$x and $y$y-values (respectively) for each characteristic.

### Application to orbits

Orbits around a planetary body are described by the conic sections (circle, ellipse, parabola and hyperbola).

 Elliptical orbits around the Sun. Not to scale.

The orbits of the planets around the Sun and the moon around the Earth are all elliptical (some are very close to circular though). The larger mass being orbited will be at one of the foci of the ellipse. For example, the Sun will be at one of the foci of the Earth's ellipse.

#### Worked example

At the closest point in its orbit (Perihelion) Mercury is $0.31$0.31 AU away from the Sun. At its furthest point in its orbit (Aphelion) it is $0.36$0.36 AU from the Sun. Find the equation of Mercury's orbit around the Sun. Give your values to two decimal places in AU.

Think: The equation of the ellipse will be in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2+y2b2=1. We can find the values of $a$a and $c$c using the geometry of the ellipse. The value of $b$b can be found given that $c^2=a^2-b^2$c2=a2b2.

From the image above we can see that the value of $a$a will be $\frac{0.31+0.36}{2}$0.31+0.362 and $c$c will be $\frac{0.36-0.31}{2}$0.360.312.

Do: Solve for $a$a and $c$c and then sub into equation for $b$b.

 $a$a $=$= $\frac{0.31+0.36}{2}$0.31+0.362​ $=$= $0.335$0.335 $c$c $=$= $\frac{0.36-0.31}{2}$0.36−0.312​ $=$= $0.025$0.025 $b$b $=$= $\sqrt{a^2-c^2}$√a2−c2 $=$= $\sqrt{0.335^2-0.025^2}$√0.3352−0.0252 $\approx$≈ $0.334$0.334 (3 d.p.)

If we create our $xy$xy plane so the major axis is aligned with $x$x-axis and the $y$y-axis is aligned with the minor axis we can determine the equation of the orbit.

$\frac{x^2}{0.335^2}+\frac{y^2}{0.334^2}=1$x20.3352+y20.3342=1

Planetary orbits

Here are a few terms that often appear in questions about orbits:

• Astronomical Units (AU): A unit equal to the distance between the Sun and the Earth, $150$150 million km or $93$93 million mi.
• Perihelion: The point in an object's orbit where it is closest to the Sun.
• Aphelion: The point in an object's orbit where it is furthest from the Sun.

Note: "Peri-" means closest point to and "apo-" means furthest point from. The next part of the word determines what is being orbited, for example, "-helion" means its orbiting the sun. Similarly "-gee" means its orbiting Earth, and "-jove"  orbiting Jupiter.

#### Practice Questions

##### QUestion 1

The equation $\frac{x^2}{324}+\frac{y^2}{289}=1$x2324+y2289=1 models the elliptical opening of a wind tunnel that is $36$36 metres wide and $34$34 metres high.

True or false?

1. True

A

False

B

True

A

False

B

##### Question 2

We want to figure out if a truck that is $8$8 feet wide carrying a load that reaches $6$6 feet above the ground will clear the semielliptical archway shown in the figure?

1. Find the equation for the bridge in terms of $y$y, the height of the bridge, $x$x metres from the centre of the archway.

Use a coordinate system with the $x$x-axis on the ground and the origin $\left(0,0\right)$(0,0) at the centre of the archway.

Give your answer in the standard form for the equation of an ellipse (rather than a semiellipse).

2. Solve for $y$y, the height of the archway $4$4 feet from the centre.

Give your answer to the nearest hundredth of a foot.

3. Will the truck clear the bridge?

Yes

A

No

B

Yes

A

No

B

##### Question 3

The Juno spacecraft entered an elliptical orbit around Jupiter in July 2016. It will perform $33$33 orbits during its mission. In 2021 the mission will be complete and the spacecraft will be deliberately driven into the atmosphere to be destroyed.

1. The formula describing Juno's orbit is $36x^2+400y^2=14400$36x2+400y2=14400, where values are given in RJ (the radius of Jupiter). What is the formula of the elliptical orbit in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2+y2b2=1?

2. Find the distance between Jupiter and Juno at its perijove. Perijove is the point in Juno's orbit where it is closest to Jupiter. Give your answer to two decimal places in RJ.

3. Find the distance between Jupiter and Juno at its apojove. Apojove is the point in Juno's orbit where it is furthest from Jupiter. Give your answer to two decimal places in RJ.

4. Kepler's third law relates the orbital period, $T$T in days, and the semi-major axis, $a$a in RJ, with the equation $a^3=2.74T^2$a3=2.74T2.

What is the orbital period of Juno? Give your answer to the nearest day.

5. If Juno completes its mission, how many days will it have spent orbiting Jupiter before entering its atmosphere?

### Outcomes

#### M8-1

Apply the geometry of conic sections

#### 91573

Apply the geometry of conic sections in solving problems