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Precalculus / Conic Sections / Parabolas

Conic Sections Reference

Parabolas as Conic Sections

A parabola is the set of all points the same distance from a focus and a directrix. This geometric definition explains the focus, directrix, focal width, and latus rectum.

Fact Table

Form Opens Focus and Directrix
$(x-h)^2=4p(y-k)$ Up if $p>0$, down if $p<0$. Focus $(h,k+p)$, directrix $y=k-p$.
$(y-k)^2=4p(x-h)$ Right if $p>0$, left if $p<0$. Focus $(h+p,k)$, directrix $x=h-p$.
Vertex Turning point. $(h,k)$
Latus rectum Segment through the focus perpendicular to the axis. Length $|4p|$.

Content Formulas

Vertical Axis
$$(x-h)^2=4p(y-k)$$
Horizontal Axis
$$(y-k)^2=4p(x-h)$$
Focal Width
$$\text{length of latus rectum}=|4p|$$
Function Connection
$$y=a(x-h)^2+k\quad\Longleftrightarrow\quad a=\frac1{4p}$$
The squared variable tells us the direction: if $x$ is squared, the parabola opens up or down; if $y$ is squared, it opens left or right.

Classic Examples

Find Focus and Directrix

Find the focus, directrix, and focal width of $(x-2)^2=12(y+1)$.

Solution StrategyMatch the equation to $(x-h)^2=4p(y-k)$.
$$(x-2)^2=12(y-(-1))$$ $$h=2,\quad k=-1,\quad 4p=12$$ $$p=3$$ $$\text{focus: }(2,2)$$ $$\text{directrix: }y=-4$$ $$\text{focal width: }12$$

Write an Equation

A parabola has vertex $(1,-2)$ and focus $(5,-2)$. Write its equation.

Solution StrategyThe focus is to the right of the vertex, so use the horizontal form.
$$h=1,\quad k=-2,\quad p=4$$ $$(y+2)^2=4(4)(x-1)$$ $$(y+2)^2=16(x-1)$$