Applications of conic sections

 

The conic sections were first discovered by Menaechaus in ancient Greece  around 350 BCE although the first detailed study was completed by Apollonius of Perga around 200BCE. Indeed it was Apollonius who gave the ellipse, the parabola and the hyperbola their modern names.

Apollonius' work greatly influenced the minds of the natural philosophers of the scientific revolution which began in the early 17th Century - philosophers like René Descartes, Johannes Kepler and Isaac Newton.

Astronomical importance

But their scientific importance was not realised until Kepler harnessed the conics to the motion of the planets. Kepler's first of three laws states that the planets move in elliptical orbits with the Sun at one focus, and this law leads directly to Isaac Newton's inverse square law of gravitational attraction.

Periodic comets also move in elliptical orbits with the Sun at one focus. Halley's comet, for example, orbiting the Sun every 75 years, tracks an elliptical path with an eccentricity of 0.967. Its path is in a plane inclined downward from the Sun 18 degrees from the plane of the planets as shown here:

http://www.davidreneke.com/catching-halleys-comet-in-2061/ 

There are other comets whose speeds are too great to be confined to a closed curve - their paths are hyperbolic, travelling toward the Sun, bending toward it, but then disappearing into oblivion never to be sighted again. The boundary between these comets and the periodic comets is the parabola.

There are also the moons of planets that travel in ellipses. Our own Moon, for example, travels along an elliptical path of eccentricity of 0.055 at an average speed of a little over a kilometre a second. 

There is also the four large moons of Jupiter - Io, Europa, Ganymede and Callisto. First spotted by Galileo in 1610 through a telescope, their almost circular dance around the great gas giant was so regular, that their relative positions were used as a clock in the sky for navigators at sea.  

Cornerstones of invention

But there are other incredible properties of the conics that have become the principle of many inventions. We all know how important the circle has been to civilisation, but properties of other conics lie at the heart of many wonderful inventions.

Here are three such inventions:

• Using the reflective property of a parabola in the design of radio telescopes
• Using the reflective property of an ellipse in a medical procedure known as Lithotripsy to pulverise kidney stones
• Estimating position using the technique of Time Difference Of Arrival using the difference in distance property of an hyperbola. 

You will find explanations of the reflective properties of the ellipse and parabola  in the links below, but there are so many others. 

Applications of ellipses

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=a2−b2.

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	$\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1	$\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)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 $\left(h-c,k\right)$(h−c,k)	$\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,k−c)
Vertices	$\left(h+a,k\right)$(h+a,k) and $\left(h-a,k\right)$(h−a,k)	$\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,k−a)
Co-vertices	$\left(h,k+b\right)$(h,k+b) and $\left(h,k-b\right)$(h,k−b)	$\left(h+b,k\right)$(h+b,k) and $\left(h-b,k\right)$(h−b,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=a2−b2.

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}$0.36−0.312​.

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

 

 

Practice Questions

QUestion 1

Question 2

Question 3

Applications of parabolic functions

As a conic section, the parabola fits between the ellipse and the hyperbola.

It is the boundary between the open and closed path of a comet as it is pulled toward the Sun.  If the comet is too slow, it becomes trapped into an elliptical orbit - too fast and it escapes along a hyperbolic trajectory.

It is the path taken by a cannon ball when fired out of a cannon at some particular angle of elevation. It is the path of a stream of water as it pushes out of a hose fitting. In fact the parabolic trajectory is a universal one - a stone thrown by someone on the surface of the moon follows a parabola.

It is the natural shape formed by the cable on the Brooklyn Bridge and the Golden Gate Bridge in San Francisco.   

The reflective property

The reflective property is perhaps the most scientifically utilised feature of the parabola. Radio telescopes and lighthouses work because of it. Torches and car headlights use parabolic mirrored surfaces that reflect light rays in such a way as to prevent the light from scattering too much. Concrete constructions of parabolic surfaces were even used in England in World War 2 to bring the reflection of distant sounds of enemy aircraft to a single point, so they could be heard.

The property

The property can be stated as follows:

Light rays emanating from the focus of a parabola will strike the inside surface (assuming its made of a mirrored compound) and be reflected back along lines parallel to the parabola's axis of symmetry.

The reflection property is illustrated in the following diagram.

The light ray shown in blue highlights the Law of Reflection. The law guarantees that the two angles $SPN$SPN and $NPQ$NPQ measured against the normal $PN$PN as the incident angle $i^{\circ}$i∘ and the reflection angle $r^{\circ}$r∘ are equal irrespective of where the ray strikes.  

 

A way to the proof    

While not a complete proof, the following investigation points the way to an understanding of why the parabola exhibits this property. We'll use an example to illustrate to concept.  

The parabola with equation given by $x^2=36y$x2=36y opens upward. Its vertex is located at the origin and the focal length can be found by solving $4a=36$4a=36. With $a=9$a=9 the focus $S$S has coordinates $\left(0,9\right)$(0,9) as shown in the diagram below.    

We can easily verify that $P$P, with coordinates $\left(12,4\right)$(12,4) is a point on the parabola. 

By calculus methods we can show that the gradient of the tangent $TPR$TPR at $P$P is given by $m=\frac{2}{3}$m=23​ and therefore the equation of the tangent becomes $y-4=\frac{2}{3}\left(x-12\right)$y−4=23​(x−12) which, when simplified, becomes $y=\frac{2}{3}x-4$y=23​x−4.

This tangent cuts the $y$y axis at $\left(0,-4\right)$(0,−4).  

The distance $ST=9+4=13$ST=9+4=13, as is the distance $SP=\sqrt{\left(12-0\right)^2+\left(4-9\right)^2}=13$SP=√(12−0)2+(4−9)2=13. This means that the triangle $SPT$SPT is isosceles with base angles equal (shown as $\theta$θ in the diagram). 

Now here's the punchline.

We know from the Law of Reflection that angles $SPN=i^{\circ}$SPN=i∘ and $NPQ=r^{\circ}$NPQ=r∘ are equal. This implies that the two adjacent angles $SPT$SPT and $QPR$QPR are also equal also. Hence, since angle $SPT$SPT is $\theta$θ, then angle $QPR$QPR is also $\theta$θ. 

But this implies that reflected ray $PQ$PQ must be parallel to the axis of the parabola because of the equal corresponding angles $QPR$QPR and $STP$STP.

This establishes that the reflective property holds for the single ray $SP$SP emanating from the focus of the parabola given by $x^2=36y$x2=36y.  

What the proof looks like  

In the above investigation, we looked at a particular parabola and a particular point $P$P. A complete proof shows that for any parabola $x^2=4ay$x2=4ay with focus $S$S and for any variable point $P$P on it, the tangent $TP$TP becomes the third side of a variable isosceles triangle $STP$STP. It thus follows that all reflected rays emanating from the focus must be parallel to the parabola's axis of symmetry.

The reverse is also true. Parallel rays which enter the parabola will strike its surface and reflect back to the focus. This is the principle behind radio telescopes, where incoming signals from outer space are collected after reflection at the focus of the parabola.

 

An application

A $70$70 metre antenna operates at the Tidbinbilla tracking station. It is the largest steerable parabolic antenna in the Southern Hemisphere. Weighing more than $3000$3000 tonnes it rotates on a film of oil. The reflector surface is made up of $1,272$1,272 aluminium panels with a total surface area of $4,180$4,180 square metres.

The photograph has been overlaid with coordinate axes together with a depiction of a cross sectional parabolic curve.

Assuming that the focal length of the parabola is $27$27 metres and the diameter of the dish is $70$70 metres, we  seek an equation for the parabola relative to the coordinate axes and the depth of the dish at the centre.

The form of the equation is given as $x^2=4ay$x2=4ay, opening from above with the vertex at the origin. Since the focal length is $27$27, the equation becomes $x^2=108y$x2=108y.

Because the diameter is $70$70 metres, the point $B$B in the photograph must have coordinates given by $\left(35,y_B\right)$(35,yB​) where the value of $y_B$yB​ can be found.

$x^2$x2	$=$=	$108y$108y
$35^2$352	$=$=	$108\left(y_B\right)$108(yB​)
$1225$1225	$=$=	$108\left(y_B\right)$108(yB​)
$\therefore$∴    $y_B$yB​	$\approx$≈	$11.343$11.343

This means the depth of the dish at the centre is $11.343$11.343 metres.

 

Practice Questions

Question 4

Question 5

Question 6

 

Applications of hyperbolas

Horizontal hyperbolas

A horizontal hyperbola is defined by the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=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=√a2+b2.
• The transverse axis, which connects the vertices. It has length $2a$2a.
• The conjugate axis, which is perpendicular to the conjugate axis with length $2b$2b.
• The asymptotes, which are the two lines that bound the hyperbola. They are defined by $y=\pm\frac{b}{a}x$y=±ba​x.

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

 

Vertical hyperbolas

A vertical hyperbola is defined by the equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2​−x2b2​=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.

 

Translated hyperbolas

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)2a2​−(y−k)2b2​=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=±ba​(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)2a2​−(x−h)2b2​=1.

 

Worked Example

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 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 $d_1>d_2$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.

 

The time delay technique

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.

 

Practice question

Question 7

Applications of circles

We have looked at the general equation of a circle, $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2, which is centred at the point $\left(h,k\right)$(h,k) and has a radius of $r$r. A special case of this equation occurs when the circle is centred at the origin. In this case, $h$h and $k$k are both $0$0 and so the equation takes the form $x^2+y^2=r^2$x2+y2=r2.

We can make use of our knowledge of circles and their equations to solve various problems involving circles. Let's take a look at a few properties.

• The equation of a circle describes all of the points that lie exactly a distance of $r$r away from the centre point $\left(h,k\right)$(h,k). Any points which are closer than $r$r units to the centre lie inside the circle, and any points that are further than $r$r units from the centre lie outside the circle.

• Two points which lie on opposite ends of a diameter of a circle have the centre of the circle as their midpoint. 

• The area of a circle of radius $r$r is given by $A=\pi r^2$A=πr2. Similarly, the length of the circumference of a circle of radius $r$r is given by $C=2\pi r$C=2πr.
• A circle can be inscribed in a square as in the image below. This square is the smallest possible square that can fit around the circle, which touches at four points and has side lengths equal to the length of the diameter of the circle.

 

Let's go through an example now.

Worked example

A circle of radius $3$3 units is centred at the point $\left(12,5\right)$(12,5). What is the shortest distance from the origin to the circle?

Think: We can start by drawing a diagram of the situation, to easily see what's going on.

From the diagram, we can see that the closest point to the origin is between the centre and the origin. In fact, if we draw a line from the origin to the centre of the circle it will intersect the circle at the closest point.

This means that the shortest distance from the origin to the circle will be equal to the distance between the origin and the centre, minus the radius of the circle.

Do: We know that the radius of the circle is $3$3 units, and we can use the distance formula to find the distance between the origin and the centre:

$d$d	$=$=	$\sqrt{\left(12-0\right)^2+\left(5-0\right)^2}$√(12−0)2+(5−0)2
	$=$=	$\sqrt{12^2+5^2}$√122+52
	$=$=	$\sqrt{144+25}$√144+25
	$=$=	$\sqrt{169}$√169
	$=$=	$13$13

Subtracting the radius from this, we can see that the shortest distance from the origin to the circle is $13-3=10$13−3=10 units.

 

Practice questions

Question 8

Question 9

Question 10