Parabolas: Practice

A parabola balances a distance to a point against a distance to a line. Writing $F$ for the focus and $d$ for the directrix, the defining condition is

$PF = \operatorname{dist}(P, d).$

The other three describe different curves. A fixed distance from a fixed point is a circle, points equidistant from two fixed points form the perpendicular bisector of the segment joining them, and points equidistant from two lines form an angle bisector.

In the form $x^2 = 4py$ the focus is $(0,p)$ and the directrix is the line $y = -p$.

$4p = 8 \;\Longrightarrow\; p = 2 \;\Longrightarrow\; y = -2$

The directrix is a line, so the answer is an equation. It sits the same distance below the vertex as the focus sits above it.

The square is on $x$, so the axis is vertical and the curve opens up or down. Read off $4p$.

$4p = -12 \;\Longrightarrow\; p = -3$

A negative $p$ puts the focus at $(0,-3)$, below the vertex, and the directrix at $y = 3$, above it. A parabola always bends toward its focus and away from its directrix, so it opens downward.

Here the square is on $y$, so the axis is horizontal and the curve opens left or right. Compare with $y^2 = 4px$.

$4p = 4 \;\Longrightarrow\; p = 1$

The focus is at $(p, 0) = (1,0)$, to the right of the vertex, and the curve bends toward its focus, so it opens to the right.

The focal width of $x^2 = 4py$ is $|4p|$, which is the size of the coefficient sitting in front of $y$, here $20$.

To see why, note $4p = 20$, so $p = 5$ and the focus is at height $y = 5$. Substituting that height into the equation gives the chord's endpoints.

$x^2 = 20(5) = 100 \;\Longrightarrow\; x = \pm 10$

The chord runs from $(-10, 5)$ to $(10, 5)$, so its length is $20$.

The focus is directly below the vertex, so the axis is vertical and $p$ is the signed distance from the vertex to the focus, $p = -1$.

$x^2 = 4py = 4(-1)y = -4y$

The distractor $x^2 = -y$ has $4p = -1$, so its focus would be at $\left(0, -\tfrac14\right)$, and $y^2 = -4x$ opens sideways, with focus $(-1, 0)$.

Distance to a line means the perpendicular distance. The line $y = -1$ is horizontal, so the perpendicular is vertical and only the $y$-coordinates matter.

$|5 - (-1)| = |6| = 6$

The $x$-coordinate $2$ plays no part: every point at height $5$ is $6$ units above the line $y = -1$.

Opening upward with a vertical axis means the form is $x^2 = 4py$, and the focus lies $p$ above the vertex, so $p = 3$.

$x^2 = 4(3)y = 12y$

The coefficient is $4p = 12$, not $p = 3$. Writing $x^2 = 3y$ would put the focus only $\tfrac34$ of a unit above the vertex.

A point is on the curve exactly when its coordinates satisfy the equation. Test $(2,1)$:

$2^2 = 4 \quad\text{and}\quad 4(1) = 4$

The two sides agree, so $(2,1)$ is on the parabola. The others fail: $(1,2)$ gives $1 \neq 8$, $(4,2)$ gives $16 \neq 8$, and $(2,4)$ gives $4 \neq 16$.

For $x^2 = 4py$ the vertex is the origin and the directrix is the line $y = -p$.

$p > 0 \;\Longrightarrow\; -p < 0$

So the directrix sits below the vertex, while the focus $(0, p)$ sits above it. The two are always on opposite sides of the vertex, at equal distances.

This is vertex form, $y = a(x - h)^2 + k$, whose vertex is $(h, k)$.

$h = 1, \qquad k = 5$

The subtraction inside the square flips the sign, so $(x-1)^2$ means $h = +1$, while the $+5$ outside is added directly. The vertex is $(1,5)$.

The axis is defined as the line through the focus perpendicular to the directrix, so the focus is always on the axis.

It cannot be on the curve. If the focus $F$ were a point of the parabola, its distance to itself would have to equal its distance to the directrix:

$0 = FF = \operatorname{dist}(F, d),$

which would put $F$ on the directrix, and the definition forbids that. It cannot be at the vertex either, since the vertex is $|p| \neq 0$ away from it.