In these notes we review conic sections. We start with a video review of the basic geometric definition, and consider the different types.
Video introduction
A parabola is the set of all points in a plane that are equidistant from a fixed point $F$, called the focus, and a fixed line $\ell$, called the directrix.
We derive a formula for a parabola with focus $F$ and directrix $\ell$ in the following video.
Video lecture
Suppose $p > 0$. Find an equation of the parabola with focus $(0,p)$ and directrix $y = -p$.
Solution
Let $P(x,y)$ represent a point on the parabola. The distance between the $P$ and the focus $F$ is
$$
d(P,F) = \sqrt{x^2 + (y-p)^2\,}.
$$
The distance between $P$ and the directrix $\ell$ is
$$
d(P,\ell) = \sqrt{(x-x)^2 + (y+p)^2\,} = \big\vert y + p \big\vert.
$$
Setting these equal and squaring both sides, yields
$$
(y+p)^2 = x^2 + (y-p)^2.
$$
Mulitplying out both sides, and canceling and combining like terms gives the equation,
$$
x^2 = 4py.
$$
Find the focus and directrix of the parabola $y^2 +10x = 0$.
Hint
Repeat the argument of the previous example for a parabola with focus $(p,0)$ and directrix $x = -p$. Then find $p$.
Check your answer
The focus is $\left(-\tfrac{5}{2},0\right)$. The directrix is $x = \tfrac{5}{2}$.
Find an equation for the parabola with focus $(a,b)$ and directrix $y = c$, where $a, b, c$ are real constants.
Hint
You could recreate the argument presented in the video lecture. Or, you could shift the parabola so that the vertex is at the origin; compare with a previous example; then shift back.
Check your answer
$$(x-a)^2 = 2(b-c)\left(y-\frac{b-c}{2}\right)$$
An ellipse is the set of all points for which the sum of the distances from two fixed points is constant. The two fixed points are called the foci (plural of focus) of the ellipse.
Video lecture
Find the foci of the ellipse
$$
\frac{x^2}{16} + \frac{y^2}{9} = 1.
$$
Check your answer
The foci are $\left(0,\pm \sqrt{7}\right)$.
A hyperbola is the set of all points for which the difference between the distances to two fixed points is constant. If $P$ is a point on the hyperbola, and $F_1$ and $F_2$ are the fixed points, then
$$
d(P,F_1) - d(P,F_2) = \text{constant}.
$$
The points $F_1$ and $F_2$ are called the foci of the hyperbola.
Video lecture
Consider the hyperbola given by the formula
$$
\dfrac{x^2}{a^2} - \frac{y^2}{b^2} = 1,
$$
where $a$ and $b$ are positive constants. Find the foci, vertices, and asymptotes of the hyperbola.
Check your answer
The foci are $(\pm c,0)$ where $c^2 = a^2 + b^2$. The vertices are $(\pm a,0)$. The asymptotes are $y = \pm \left(\tfrac{b}{a}\right) x$.
Video solution
Questions? You can ask them below.
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