Identifying and writing 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 centre or vertex
• The location of foci
• The equations of directrices
• Particular points on the conic, including axes intercepts

Lets look at a few examples:

Example 1 

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

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 is the parabola's focal length. 

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.

 

Example 2

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

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 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1 with $b>a$b>a.  

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

This means we know that the equation has the form $\frac{x^2}{a^2}+\frac{y^2}{24^2}=1$x2a2​+y2242​=1, and the only thing left to do is to find $a$a. We do this by substitution into the eccentricity formula $e=\frac{\sqrt{b^2-a^2}}{b}$e=√b2−a2b​ (Note here that the length $b$b, greater than $a$a, is in the denominator).

$e$e	$=$=	$\frac{\sqrt{b^2-a^2}}{b}$√b2−a2b​
$\frac{1}{2}$12​	$=$=	$\frac{\sqrt{24^2-a^2}}{24}$√242−a224​
$12$12	$=$=	$\sqrt{576-a^2}$√576−a2
$144$144	$=$=	$576-a^2$576−a2
$\therefore$∴    $a^2$a2	$=$=	$432$432
$a$a	$=$=	$12\sqrt{3}$12√3

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

Example 3

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

The eccentricity of $5$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 $ae=10$ae=10. 

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. 

More Worked Examples

Question 1

Question 2

Question 3