9.06 Eccentricity and identifying conic sections

Identify and graph conic sections 

The circle, ellipse, parabola and hyperbola are all types of conic sections. When a plane intersects a cone at different angles it will produces one of the four curves depending on the slope of the cone and of the plane.

The four conic sections found by intersecting a cone.

The standard form(s) of each equation (for both the horizontal and vertical orientations) and the important graphing features for each conic section are presented in the table below:

Conic section	Equation	Graph	Characteristics
Ellipse	$\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1		Center: $\left(h,k\right)$(h,k) Vertices: $\left(h\pm a,k\right)$(h±a,k) Co-vertices: $\left(h,k\pm b\right)$(h,k±b)
	$\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1		Center: $\left(h,k\right)$(h,k) Co-vertices: $\left(h\pm b,k\right)$(h±b,k) Vertices: $\left(h,k\pm a\right)$(h,k±a)
Circle	$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2		Center: $\left(h,k\right)$(h,k) Radius: $r$r
Hyperbola	$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1		Center: $\left(h,k\right)$(h,k) Vertices: $\left(h\pm a,k\right)$(h±a,k) Asymptotes: $y=\pm\frac{b}{a}\left(x-h\right)+k$y=±ba​(x−h)+k
	$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1		Center: $\left(h,k\right)$(h,k) Vertices: $\left(h,k\pm a\right)$(h,k±a) Asymptotes: $y=\pm\frac{a}{b}\left(x-h\right)+k$y=±ab​(x−h)+k
Parabola	$y=a\left(x-h\right)^2+k$y=a(x−h)2+k		Vertex: $\left(h,k\right)$(h,k) $a$a determines direction of opening $y$y-intercept: when $x=0$x=0 $x$x-intercepts: when $y=0$y=0
	$x=a\left(y-k\right)^2+h$x=a(y−k)2+h		Vertex:$\left(h,k\right)$(h,k) $a$a determines direction of opening $y$y-intercepts: when $x=0$x=0 $x$x-intercept: when $y=0$y=0

 

The general conic form

Each conic section has its own standard form that makes it easiest to read off its characteristics. There is, however, a general form for all conic sections:

$Ax^2+By^2+Cx+Dy+E=0$Ax2+By2+Cx+Dy+E=0,

where $A$A, $B$B, $C$C, $D$D and $E$E are real numbers. All conic sections can be represented by this equation, but not all choices of $A$A, $B$B, $C$C, $D$D and $E$E will result in a conic section - as we will see below. This general form is useful for classifying which type of conic section a particular equation represents.

If an equation in this form does represent a conic section, we can determine which one it will be by looking at the values of $A$A and $B$B. The table below outlines this:

	Conic		Value of $A$A and $B$B
Parabola			$A=0$A=0 or $B=0$B=0
Ellipse			$A$A and $B$B have the same sign, that is, $A\times B>0$A×B>0
Circle			$A$A = $B$B
Hyperbola			$A$A and $B$B have the opposite sign, that is, $A\times B<0$A×B<0

 

To graph a conic in this form it is easiest to rearrange the equation (which often involves completing the square) into a form where it is easier to find its identifying features (center, vertices, radius etc.).

 

Exploration

Use the applet below to investigate the different curves represented by the general conic equation, by varying just the values of $A$A and $B$B.

• Can you find all four of the conic sections? Look at the relationship between $A$A and $B$B for each type.
• Can you find conditions under which the graph forms a line (or two lines)? Think about how this relates to planes intersecting a cone as in the image at the top.

Note: The values of $C$C, $D$D and $E$E are fixed in the applet at $-10$−10, $10$10 and $10$10.

 

Worked example

Question 1

The following equations describe different conic sections. Determine the conic section represented:

• $3x^2-2x+\frac{y^2}{4}+y-10=0$3x2−2x+y24​+y−10=0
• $-2x^2+7x+9y-2y^2+4=0$−2x2+7x+9y−2y2+4=0
• $x^2+10x-3y=0$x2+10x−3y=0
• $5y^2+6x=5x^2-9y$5y2+6x=5x2−9y

Think: Since we already know that these equations represent conics, we can use the values of $A$A and $B$B to determine which conics they are. We can rearrange each equation so that it is in the form $Ax^2+By^2+Cx+Dy+E=0$Ax2+By2+Cx+Dy+E=0 and then read off the values.

Do:

• This equation doesn't need to be rearranged, $A=3$A=3 and $B=\frac{1}{4}$B=14​. They are the same sign (both positive) but not equal, so the conic is an ellipse.
• The coefficients are $A=-2$A=−2 and $B=-2$B=−2. They are the same sign (both negative) and equal, so the conic is a circle.
• The coefficients are $A=1$A=1 and $B=0$B=0. $A$A or $B$B is zero, so the conic is a parabola.
• We need to rearrange this equation first to the form $5y^2+6x-5x^2+9y=0$5y2+6x−5x2+9y=0. So $A=-5$A=−5 and $B=5$B=5. The signs are opposite, so the conic is a hyperbola.

Reflect: We could only use this technique because the questions told us these equations definitely did describe conics. Otherwise we would have had to rearrange the equation to make sure they had a solution. Note also that if $A$A and $B$B have the same sign, it will be an ellipse, but if they are equal as well a circle is a more accurate description.

 

Question 2

Determine the curve produced by the equation $3x^2+5y^2-6x-20y+8=0$3x2+5y2−6x−20y+8=0 and draw the graph.

Think: $A$A and $B$B are both positive so this looks like it will be an ellipse. We should first try to rearrange it into 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 to make sure it is an ellipse. We can then use the values of $a$a, $b$b, $h$h and $k$k to help draw the graph (if it is an ellipse).

Do: Rearrange the equation into the desired form by completing the square

$3x^2+5y^2-6x-20y+8$3x2+5y2−6x−20y+8	$=$=	$0$0
$3\left(x^2-2x\right)+5\left(y^2-4y\right)$3(x2−2x)+5(y2−4y)	$=$=	$-8$−8
$3\left(\left(x-1\right)^2-1\right)+5\left(\left(y-2\right)^2-4\right)$3((x−1)2−1)+5((y−2)2−4)	$=$=	$-8$−8
$3\left(x-1\right)^2+5\left(y-2\right)^2$3(x−1)2+5(y−2)2	$=$=	$3+20-8$3+20−8
$\frac{3\left(x-1\right)^2}{15}+\frac{5\left(y-2\right)^2}{15}$3(x−1)215​+5(y−2)215​	$=$=	$\frac{15}{15}$1515​
$\frac{\left(x-1\right)^2}{5}+\frac{\left(y-2\right)^2}{3}$(x−1)25​+(y−2)23​	$=$=	$1$1

The equation describes a ellipse where $a=\sqrt{5}$a=√5 and $b=\sqrt{3}$b=√3, with a center at $\left(1,2\right)$(1,2).

Notice the vertices at $\left(1\pm\sqrt{5},2\right)$(1±√5,2) and co-vertices at $\left(1,2\pm\sqrt{3}\right)$(1,2±√3).

 

Question 3

What does the following equation represent $2x^2-3y^2-12x+30y-57=0$2x2−3y2−12x+30y−57=0? If it is a conic section, draw a graph of the conic.

Think: The coefficients of $x^2$x2 and $y^2$y2 have opposite signs, so if this is a conic section then it will be a hyperbola. We can try to rearrange the equation into the more useful hyperbola form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1.

Do:

$2x^2-3y^2-12x+30y-57$2x2−3y2−12x+30y−57	$=$=	$0$0
$2\left(x^2-6x\right)-3\left(y^2-10\right)$2(x2−6x)−3(y2−10)	$=$=	$57$57
$2\left(\left(x-3\right)^2-9\right)-3\left(\left(y-5\right)^2-25\right)$2((x−3)2−9)−3((y−5)2−25)	$=$=	$57$57
$2\left(x-3\right)^2-18-3\left(y-5\right)^2+75$2(x−3)2−18−3(y−5)2+75	$=$=	$57$57
$2\left(x-3\right)^2-3\left(y-5\right)^2$2(x−3)2−3(y−5)2	$=$=	$57+18-75$57+18−75
$2\left(x-3\right)^2-3\left(y-5\right)^2$2(x−3)2−3(y−5)2	$=$=	$0$0
$\left(y-5\right)^2$(y−5)2	$=$=	$\frac{2}{3}\left(x-3\right)^2$23​(x−3)2
$y$y	$=$=	$\pm\sqrt{\frac{2}{3}}\left(x-3\right)+5$±√23​(x−3)+5

 

Reflect: The equation looks like it would be a hyperbola, however when we completed the squares there was no constant term. The result is just a pair of intersecting lines.

 

Practice questions

Question 4

Question 5

Question 6

 

Eccentricity of a conic section

Every conic section has what is called an eccentricity. The eccentricity is a value $e>0$e>0 (not Euler's number), which characterizes the shape of the conic.

Identifying conics from the eccentricity

Given enough information about a conic, it is possible to find its eccentricity. Also, given the eccentricity, we can get some information about the curve.

In particular, 

Eccentricity	Type of Conic	Calculation of Eccentricity
$e=0$e=0	Circle	$0$0
$00<e<1	Ellipse	$\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{a^2-b^2}}{a}$√1−b2a2​=√a2−b2a​ or $\frac{c}{a}$ca​, where $c=\sqrt{a^2-b^2}$c=√a2−b2
$e=1$e=1	Parabola	$1$1
$e>1$e>1	Hyperbola	$\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{a^2+b^2}}{a}$√1+b2a2​=√a2+b2a​ or $\frac{c}{a}$ca​, where $c=\sqrt{a^2+b^2}$c=√a2+b2

 

The ellipse

These two ellipses have the same eccentricity:

They are both stretched the same amount but the direction of stretch is different. Note that if we call the length of the semi-major axis $sM$sM and the length of the semi-minor axis $sm$sm, then for both ellipses:

$e=\frac{\sqrt{\left(sM\right)^2-\left(sm\right)^2}}{sM}$e=√(sM)2−(sm)2sM​

Clearly we would be in all sorts of trouble if, under the square root, a larger square was being subtracted from a smaller square.

The hyperbola

The standard formula for a horizontal hyperbola is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1, and its eccentricity is given by $e=\frac{\sqrt{a^2+b^2}}{a}$e=√a2+b2a​. Note the similarity to the ellipse, except the signs are flipped around in both formulae.  

The two possibilities are depicted here:

 

Finding the Eccentricity

Note that to find $e$e, we either need $a$a and $b$b or we need enough information to find $a$a and $b$b.

The information may be given algebraically, or graphically, or simply stated in a sentence. We will demonstrate with a few examples:

Worked examples

Question 7

Find the eccentricity of the conic defined by $9x^2+100y^2=900$9x2+100y2=900

Think: To know which formula to use, we need to determine which type of conic this is. We recognize this is an ellipse, so now we need to rearrange to standard form to determine which of the two ellipse formulas to use.

Do: 

$9x^2+100y^2$9x2+100y2	$=$=	$900$900
$\frac{x^2}{100}+\frac{y^2}{9}$x2100​+y29​	$=$=	$1$1

This means $a=10$a=10 and $b=3$b=3, so the ellipse is elongated horizontally. We will use $\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{a^2-b^2}}{a}$√1−b2a2​=√a2−b2a​.

$e$e	$=$=	$\frac{\sqrt{a^2-b^2}}{a}$√a2−b2a​	State the formula
	$=$=	$\frac{\sqrt{10^2-3^2}}{10}$√102−3210​	Fill in given information
	$=$=	$\frac{\sqrt{100-9}}{10}$√100−910​	Simplify
	$=$=	$\frac{\sqrt{91}}{10}$√9110​	Simplify further

 

Question 8

Calculate the eccentricity of $4y^2-16x^2=256$4y2−16x2=256.

Think: This is the equation of a hyperbola, and by rearranging it into standard form can determine if it is a horizontal or vertical hyperbola.

Do:

$4y^2-16x^2$4y2−16x2	$=$=	$256$256
$\frac{y^2}{64}-\frac{x^2}{16}$y264​−x216​	$=$=	$1$1

We can see that this is a vertical or conjugate hyperbola, therefore we will use $e=\frac{\sqrt{b^2+a^2}}{b}$e=√b2+a2b​ and we can also see that $a=4$a=4 and $b=8$b=8 (check this).

$e$e	$=$=	$\frac{\sqrt{b^2+a^2}}{b}$√b2+a2b​	State the equation
	$=$=	$\frac{\sqrt{8^2+4^2}}{8}$√82+428​	Fill in given information
	$=$=	$\frac{\sqrt{64+16}}{8}$√64+168​	Simplify
	$=$=	$\frac{\sqrt{90}}{8}$√908​	Simplify further
	$=$=	$\frac{\sqrt{5}}{2}$√52​	Simplify the radical and fraction

 

Question 9

A certain central conic of the form $\frac{x^2}{m^2}+\frac{y^2}{n^2}=1$x2m2​+y2n2​=1  passes through the point $\left(4,6\right)$(4,6) and has a semi-major axis length of $10$10. Show that, without a diagram, there are two possibilities for the eccentricity.

Think: Based on the information given, and without a diagram, we are not sure whether we are looking at an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1 or $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2​+y2a2​=1, where $a>b$a>b.

So there are in fact two possibilities for the equation, both ellipses, given as $\frac{x^2}{10^2}+\frac{y^2}{b^2}=1$x2102​+y2b2​=1 (elongated horizontally) or $\frac{x^2}{b^2}+\frac{y^2}{10^2}=1$x2b2​+y2102​=1 (elongated vertically).

Do:

Taking the first option, $\frac{x^2}{100}+\frac{y^2}{b^2}=1$x2100​+y2b2​=1,  and knowing that $\left(4,6\right)$(4,6) is a point on the conic, we have:

$\frac{4^2}{100}+\frac{6^2}{b^2}$42100​+62b2​	$=$=	$1$1
$\frac{6^2}{b^2}$62b2​	$=$=	$1-\frac{4^2}{100}$1−42100​
$\frac{6^2}{b^2}$62b2​	$=$=	$\frac{84}{100}$84100​
$\therefore$∴     $b^2$b2	$=$=	$\frac{300}{7}$3007​

This means that the ellipse could have the equation $\frac{x^2}{100}+\frac{7y^2}{300}=1$x2100​+7y2300​=1, with $a=10$a=10 and $b=\frac{10\sqrt{3}}{\sqrt{7}}$b=10√3√7​. The eccentricity would be given by $e=\frac{\sqrt{100-\frac{300}{7}}}{10}$e=√100−3007​10​ which when simplified becomes $e=\frac{2}{\sqrt{7}}$e=2√7​. 

On the other hand, the ellipse could have the equation $\frac{x^2}{b^2}+\frac{y^2}{10^2}=1$x2b2​+y2102​=1, and so proceeding in a similar manner with the point $\left(4,6\right)$(4,6), we have:

$\frac{16}{b^2}+\frac{36}{100}$16b2​+36100​	$=$=	$1$1
$\frac{16}{b^2}$16b2​	$=$=	$1-\frac{36}{100}$1−36100​
$\frac{16}{b^2}$16b2​	$=$=	$\frac{64}{100}$64100​
$\therefore$∴     $b^2$b2	$=$=	$25$25

The ellipse would then have the equation $\frac{x^2}{25}+\frac{y^2}{100}=1$x225​+y2100​=1, with $b=5$b=5 and $a=10$a=10. The eccentricity would then be given by $e=\frac{\sqrt{100-25}}{10}=\frac{\sqrt{75}}{10}$e=√100−2510​=√7510​, simplifying to $e=\frac{\sqrt{3}}{2}$e=√32​.

Reflect: We now use graphing technology to confirm this.

 

Practice questions

Question 10

Question 11

Question 12

 

Identify and write the equation of a conic section 

The equation of a conic can be found if sufficient identifying information is known. This information might include, among other things:

• The conics eccentricity (identifying the type of conic)
• The location of the conic's center or vertex
• The location of foci
• The equations of directrices
• Particular points on the conic, including axes intercepts

Lets look at a few examples:

Worked examples

question 13

Find the equation of the conic with vertex at the origin, a focus at $\left(-3,0\right)$(−3,0) and an eccentricity of $1$1

Think: A eccentricity of $1$1 immediately identifies the conic as a parabola. The focus at $\left(-3,0\right)$(−3,0) and a vertex at the origin clearly means that the parabola opens to the left, and thus must have the equation $y^2=-4ax$y2=−4ax, where $a$a is the parabola's focal length. 

Do: This means that $4a=3$4a=3 and thus $a=\frac{3}{4}$a=34​, and so the equation becomes:

$y^2$y2	$=$=	$4ax$4ax
	$=$=	$4\left(\frac{3}{4}\right)x$4(34​)x
$\therefore$∴    $y^2$y2	$=$=	$3x$3x

 The equation of the conic is thus $y^2=3x$y2=3x.

 

Question 14

Find the equation of the the conic with a center at the origin, a focus at  $\left(0,12\right)$(0,12) and an eccentricity of $\frac{1}{2}$12​.

Think: The eccentricity of $\frac{1}{2}$12​ immediately tells us that the conic is an ellipse. Note that the given focus also tells us that the major axis lies along the $y$y axis. Thus the equation of the conic has the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2​+y2a2​=1.  

Do: Since this means that the foci are given generally as $\left(0,\pm ae\right)$(0,±ae), then we know that $ae=12$ae=12, where $a$a is the length of the semi-major axis. Hence $a\left(\frac{1}{2}\right)=12$a(12​)=12 and thus  $a=24$a=24. 

This means we know that the equation has the form $\frac{x^2}{b^2}+\frac{y^2}{24^2}=1$x2b2​+y2242​=1, and the only thing left to do is to find $a$a. We do this by substitution into the eccentricity formula $e=\frac{c}{a}=\frac{\sqrt{a^2-b^2}}{a}$e=ca​=√a2−b2a​.

$e$e	$=$=	$\frac{\sqrt{a^2-b^2}}{a}$√a2−b2a​
$\frac{1}{2}$12​	$=$=	$\frac{\sqrt{24^2-b^2}}{24}$√242−b224​
$12$12	$=$=	$\sqrt{576-b^2}$√576−b2
$144$144	$=$=	$576-b^2$576−b2
$\therefore$∴    $b^2$b2	$=$=	$432$432
$b$b	$=$=	$12\sqrt{3}$12√3

The equation is thus $\frac{x^2}{432}+\frac{y^2}{576}=1$x2432​+y2576​=1.   

Question 15

Identify the conic with a center at the origin, a focus at $\left(10,0\right)$(10,0)  and an eccentricity of $5$5.

Think: The eccentricity of $e=5$e=5 means that we are dealing with an hyperbola. The focus lies along the $x$x axis, so we are looking with an equation of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1, with $c=10$c=10, so $5=\frac{10}{a}$5=10a​.

Do: This means that, since $e=5$e=5, $a=2$a=2 the equation becomes $\frac{x^2}{4}-\frac{y^2}{b^2}=1$x24​−y2b2​=1. 

For the hyperbola we have that $e=\frac{\sqrt{a^2+b^2}}{a}$e=√a2+b2a​ (note the plus sign under the square root).

$e$e	$=$=	$\frac{\sqrt{2^2+b^2}}{2}$√22+b22​
$10$10	$=$=	$\sqrt{4+b^2}$√4+b2
$100$100	$=$=	$4+b^2$4+b2
$b^2$b2	$=$=	$96$96
$\therefore$∴    $b$b	$=$=	$4\sqrt{6}$4√6

The equation is thus $\frac{x^2}{4}-\frac{y^2}{96}=1$x24​−y296​=1. 

 

Practice questions

Question 16

Question 17

Question 18