Identifying and graphing 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		Centre: $\left(h,k\right)$(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		Centre: $\left(h,k\right)$(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		Centre: $\left(h,k\right)$(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		Centre: $\left(h,k\right)$(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		Centre: $\left(h,k\right)$(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 concavity $y$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 concavity $y$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 (centre, vertices, radius etc.).

 

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 examples

Example 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.

 

Example 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 centre 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).

 

Example 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 1

Question 2

Question 3