NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Identifying and writing the equation of a conic section
Lesson

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=b2a2b (Note here that the length $b$b, greater than $a$a, is in the denominator).

$e$e $=$= $\frac{\sqrt{b^2-a^2}}{b}$b2a2b
$\frac{1}{2}$12 $=$= $\frac{\sqrt{24^2-a^2}}{24}$242a224
$12$12 $=$= $\sqrt{576-a^2}$576a2
$144$144 $=$= $576-a^2$576a2
$\therefore$    $a^2$a2 $=$= $432$432
$a$a $=$= $12\sqrt{3}$123

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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2y2b2=1, with $ae=10$ae=10

This means that, since $e=5$e=5, $a=2$a=2 the equation becomes $\frac{x^2}{4}-\frac{y^2}{b^2}=1$x24y2b2=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}$46
     

The equation is thus $\frac{x^2}{4}-\frac{y^2}{96}=1$x24y296=1

More Worked Examples

Question 1

Find the equation of the conic with a vertex at the origin, a focus at $\left(0,-2\right)$(0,2) and an eccentricity of $1$1.

Question 2

Consider the conic with a centre at the origin, a focus at $\left(4,0\right)$(4,0) and an eccentricity of $\frac{1}{3}$13.

  1. Find the value of $b^2$b2.

  2. Hence find the equation of the conic.

Question 3

Consider the conic with a centre at the origin, $y$y-intercepts at $\left(0,\pm8\right)$(0,±8) and an eccentricity of $\frac{3}{2}$32.

  1. Find the value of $b^2$b2.

  2. Hence find the equation of the conic.

 

Outcomes

M8-1

Apply the geometry of conic sections

91573

Apply the geometry of conic sections in solving problems

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