Classifying conic sections

 

Generating a conic section

A conic section (often just called a conic) is a curve generated by the intersection of the surface of a cone with a plane.

As this diagram shows, depending on the angle that the cutting plane makes with the cone, one of four curves are obtained. 

• If the plane cuts the cone horizontally, the closed curve generated becomes a circle (shown in red)
• If the cutting plane is tilted at an angle, and cuts through to make a closed curve, that curve becomes an ellipse (shown in green).
• If the cutting plane is tilted so that the major axis of the curve generated (shown as a dotted line through the blue curve) is parallel to the slant edge of the cone, then the curve becomes a parabola.
• If the cutting plane is tilted further than that required to generate a parabola, then the curve becomes a hyperbola (shown in orange). 

In the case of the hyperbola, an inverted cone balanced upside with its apex touching the apex of the first cone is often shown in the diagram (the pair of cones are referred to as a double napped cone). The plane will cut both cones and generate two hyperbolic branches. These branches will be reflections of each other if the cutting plane becomes vertical. 

 

The Greek mathematician Appolonius of Perga around 200 BCE investigated these curves extensively, but it was not until the  17th Century with the emergence of the scientific revolution that natural philosophers such as Johannes Kepler and Isaac Newton realised the physical significance of the four sections. For example, the path of a water stream from a hose is parabolic, the path of a planet is elliptical and the path of a fast comet that sweeps by our Sun once will curve around it as a hyperbola. 

This neat animated applet shows the various cuts creating the different conics.

The conics as loci    

A conic section can also be established as the locus of points. 

Imagine walking between a tree that lies close to a river's edge in such a way as to keep an equal distance between both:

 

In mathematical terms we might describe the locus as all points $P$P such that the distance from the base of the tree $S$S to $P$P is always the same distance from $P$P to the bank of the river $M$M. In other words, $SP=PM$SP=PM. 

Remarkably, we can show that this path is a parabola.   

If we changed the locus so that $SP=2PM$SP=2PM (a path closer to the river than the tree), the path changes to a hyperbola. In fact this will happen if $SP=ePM$SP=ePM where $e>1$e>1.

If we changed the locus so that $SP=SP=\frac{1}{2}PM$SP=SP=12​PM (a path closer to the tree than the river) the path becomes an ellipse - a closed loop around the tree! Again, this will always happen if $SP=ePM$SP=ePM and $e<1$e<1, shown here:

Overlaying a coordinate system

To discover the many properties of conic sections, it is convenient to overlay a cartesian coordinate system in such a way as to simplify the mathematics that arises.

To this end, we place the coordinate axes so as to make the lines of symmetry of the conic parallel and perpendicular to these axes, as shown here using three examples:

For the ellipse and the hyperbola, the red dot in the picture represents the centre of the conic, and in the parabola the red dot locates its vertex.

The equation of a conic

Now when we set up the axes in this way, the equation of every conic (including the circle) has the form:

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

with $A,B,C,D$A,B,C,D and $E$E are coefficients of each of the terms in the equation. 

To be a conic, either $A$A or $B$B or both must be non-zero (note that if $A=B=0$A=B=0 the conic would degenerate into the straight line $Cx+Dy+E=0$Cx+Dy+E=0). 

In fact we can classify the type of conic based on $A$A and $B$B:

• If the conic is an ellipse, then $AB$AB is positive
• An ellipse that has $A=B$A=B is a circle
• If the conic is a hyperbola, then $AB$AB is negative
• If the conic is a parabola, then either $A$A or $B$B (but not both) is zero. 

Unfortunately, these statements don't necessarily work the other way around (See example 4 below). 

If either $C$C or $D$D or both are non-zero, then the centre of the hyperbola or ellipse will be located away from the origin. The same applies to the parabola - if $C$C or $D$D or both are non-zero, the vertex will be away from the origin.

As for the constant coefficient $E$E, there are certain circumstances where degenerate forms can arise but these are best understood by considering a few examples.

With the axes set up like this, the equation of the hyperbola and the ellipse can always be rearranged into the standard forms $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=\pm1$(x−h)2a2​−(y−k)2b2​=±1 and $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1 respectively where $\left(h,k\right)$(h,k) locates the centre.

Likewise, the equation of the parabola can always be rearranged into the vertex form $y-k=a\left(x-h\right)^2$y−k=a(x−h)2 where the vertex is located at $\left(h,k\right)$(h,k).  

Some examples

Example 1

What type of conic is given by the equation $x^2+y^2-6x+4y-12=0$x2+y2−6x+4y−12=0. Where is the conic's centre?

Since $A=B=1$A=B=1, the conic is a circle. By completing squares on $x$x and $y$y, we can locate the centre:

$x^2+y^2-6x+4y-12$x2+y2−6x+4y−12	$=$=	$0$0
$x^2-6x+y^2+4y$x2−6x+y2+4y	$=$=	$12$12
$x^2-6x+9+y^2+4y+4$x2−6x+9+y2+4y+4	$=$=	$12+9+4$12+9+4
$\left(x-3\right)^2+\left(y+2\right)^2$(x−3)2+(y+2)2	$=$=	$25$25

This means that the circle's centre is located at $\left(3,-2\right)$(3,−2) and the circle's radius is $5$5.

Example 2

What type of conic is given by the equation $x^2-6x-3y-12=0$x2−6x−3y−12=0. Where is the conic's centre?

Since $B=0$B=0, the equation is that for a parabola. Again, completing squares:

$x^2-6x-3y-12$x2−6x−3y−12	$=$=	$0$0
$x^2-6x+9-3y-21$x2−6x+9−3y−21	$=$=	$0$0
$\left(x-3\right)^2-3\left(y+7\right)$(x−3)2−3(y+7)	$=$=	$0$0
$y+7$y+7	$=$=	$\frac{1}{3}\left(x-3\right)^2$13​(x−3)2

Hence the parabola opens upward with vertex $\left(3,-7\right)$(3,−7). 

Example 3

What type of conic is given by the equation $9x^2+4y^2+18x-16y-11=0$9x2+4y2+18x−16y−11=0? Where is the conic's centre, and find its $x$x intercepts?

Since $A\ne B$A≠B and $AB$AB is positive, the equation represents an ellipse. 

The $x$x intercepts are found by setting $y=0$y=0 and thus solving for $x$x in $9x^2+18x-11=0$9x2+18x−11=0. By using the quadratic formula, we find that $x=-1\pm\frac{2\sqrt{5}}{3}$x=−1±2√53​, and these are the $x$x intercepts.

To find the centre we again complete squares:

$9x^2+4y^2+18x-16y-11$9x2+4y2+18x−16y−11	$=$=	$0$0
$9x^2+18x+4y^2-16y$9x2+18x+4y2−16y	$=$=	$11$11
$9\left(x^2+2x+1\right)+4\left(y^2-4y+4\right)$9(x2+2x+1)+4(y2−4y+4)	$=$=	$36$36
$9\left(x+1\right)^2+4\left(y-2\right)^2$9(x+1)2+4(y−2)2	$=$=	$36$36
$\frac{\left(x+1\right)^2}{4}+\frac{\left(y-2\right)^2}{9}$(x+1)24​+(y−2)29​	$=$=	$1$1

Hence the centre of the ellipse is located at $\left(-1,2\right)$(−1,2). 

Example 4

Is $2x^2-3y^2-12x+30y-57=0$2x2−3y2−12x+30y−57=0 a hyperbola?

Since $AB=2\times-3=-6$AB=2×−3=−6, the conic looks to be a hyperbola. However, when we complete squares, we find that the equation becomes $2\left(x-3\right)^2-3\left(y-5\right)^2=0$2(x−3)2−3(y−5)2=0.

Without going into the algebra, the left hand side can be factorised into the difference of two squares and each factor can be brought to zero, so that the equation represents two straight lines with equations given by $\sqrt{2}\left(x-3\right)-\sqrt{3}\left(y-5\right)=0$√2(x−3)−√3(y−5)=0 and $\sqrt{2}\left(x-3\right)+\sqrt{3}\left(y-5\right)=0$√2(x−3)+√3(y−5)=0 as shown here:

Thus it is a necessary, but not sufficient condition that $AB$AB to be negative for the equation to represent a hyperbola.

Mathspace Worked Examples

Question 1

Question 2

Question 3