The equation of a conic can be found if sufficient identifying information is known. This information might include, among other things:
Lets look at a few examples:
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.
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 .
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.
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.
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.
Find the value of $b^2$b2.
Hence find the equation of the conic.
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.
Find the value of $b^2$b2.
Hence find the equation of the conic.
Apply the geometry of conic sections
Apply the geometry of conic sections in solving problems