Parabolas: Practice

This is the definition of a parabola with focus $(4,0)$ and directrix $x = -4$. Write the distance condition for a point $(x,y)$ and square it, which is reversible because both sides are distances.

$\sqrt{(x-4)^2 + y^2} = |x + 4|$

$x^2 - 8x + 16 + y^2 = x^2 + 8x + 16$

The $x^2$ and the constants cancel, leaving

$y^2 = 16x.$

The directrix is vertical, so the axis is horizontal and $y$ is the squared variable. Here $4p = 16$, so $p = 4$, matching the focus at $(4,0)$.

Use the definition instead of the distance formula. Since $4p = 12$, we have $p = 3$, so the directrix is $y = -3$. For a point on the curve, the distance to the focus equals the distance to the directrix.

$PF = y + 3$

Set that equal to $8$:

$y + 3 = 8 \;\Longrightarrow\; y = 5$

Check it directly: at $y = 5$, $x^2 = 60$, and the distance from $(\sqrt{60}, 5)$ to the focus $(0,3)$ is $\sqrt{60 + 4} = 8$.

Here $4p = 4$, so $p = 1$, the focus is $(0,1)$, and the directrix is $y = -1$. For any point on the curve the focal distance equals the distance to the directrix.

$PF = y + 1$

On the parabola $x^2 = 4y$ every point has $y = \tfrac{x^2}{4} \ge 0$, so $y + 1$ is smallest when $y = 0$.

$PF_{\min} = 0 + 1 = 1$

The closest point is the vertex, and its distance to the focus is $|p| = 1$. No point of a parabola is ever nearer its focus than the vertex is.

Rewrite each as $x^2 = 4py$; the focal width is then $|4p|$, the coefficient of $y$.

$y = \tfrac{x^2}{4} \Rightarrow x^2 = 4y, \quad y = \tfrac{x^2}{2} \Rightarrow x^2 = 2y, \quad y = x^2 \Rightarrow x^2 = y, \quad y = 2x^2 \Rightarrow x^2 = \tfrac12 y$

The widths are $4$, $2$, $1$, and $\tfrac12$, so $y = \tfrac{x^2}{4}$ is widest.

In general the focal width is $\left|\tfrac{1}{a}\right|$, so the smaller the leading coefficient, the wider and shallower the curve, and the farther the focus sits from the vertex.

A vertical axis with vertex $(3,-2)$ means $(x-3)^2 = 4p(y+2)$. Substitute the point $(5,2)$.

$(5-3)^2 = 4p(2+2) \;\Longrightarrow\; 4 = 16p \;\Longrightarrow\; p = \tfrac14$

The focus is $p$ above the vertex.

$\left(3, \; -2 + \tfrac14\right) = \left(3, -\tfrac74\right)$

The distractor $-\tfrac94$ is the directrix's height, $k - p$, which lies the same distance on the other side of the vertex.

Put the vertex at the origin with the dish opening upward, so the cross-section is $x^2 = 4py$. Being $24$ inches across means the rim is $12$ inches either side of the axis, and $4$ inches deep puts the rim at height $4$, so the point $(12, 4)$ is on the curve.

$12^2 = 4p(4) \;\Longrightarrow\; 144 = 16p \;\Longrightarrow\; p = 9$

The receiver belongs at the focus, which is $p = 9$ inches from the vertex. Note that the focus is well outside the $4$-inch bowl, which is normal for a shallow dish. Using the full width $24$ instead of the half-width $12$ is the usual mistake.

Substitute the point into the equation.

$(-6)^2 = 4p(3) \;\Longrightarrow\; 36 = 12p \;\Longrightarrow\; p = 3$

The square is on $y$, so the axis is horizontal and the directrix is the vertical line $x = -p$.

$x = -3$

Pairing $p = 3$ with the directrix $y = -3$ gets $p$ right but quotes a horizontal line, which cannot be the directrix of a sideways parabola: the directrix is always perpendicular to the axis.

Substitute the line into the parabola and collect terms.

$x^2 = 4(x - 1) \;\Longrightarrow\; x^2 - 4x + 4 = 0$

The left side is a perfect square, $(x-2)^2 = 0$, so the discriminant is zero and $x = 2$ is a double root.

$x = 2 \;\Longrightarrow\; y = 1$

There is exactly one shared point, $(2,1)$. A repeated root means the line touches the curve without crossing it, so this line is tangent to the parabola.

Call that foot $D$. The definition of the parabola says $PF = PD$, so $P$ is equidistant from the two points $F$ and $D$, which places $P$ on the perpendicular bisector of $FD$.

$PF = PD \;\Longrightarrow\; P \in \text{perpendicular bisector of } FD$

Every other point $Q$ of that bisector satisfies $\operatorname{dist}(Q, d) < QD = QF$, so it lies outside the curve. The line therefore touches the parabola at $P$ alone, which is what makes it the tangent, and it is exactly the mirror that swaps $F$ and $D$. That is why a ray arriving along $PD$ reflects along $PF$.

Test each point by substituting into $y^2 = 8x$.

$(2,4)\colon\; 4^2 = 16 = 8(2) \;\checkmark \qquad (8,8)\colon\; 8^2 = 64 = 8(8) \;\checkmark \qquad (0,0)\colon\; 0 = 0 \;\checkmark$

Now $(4,6)$:

$6^2 = 36 \quad\text{but}\quad 8(4) = 32$

Since $36 \neq 32$, the point $(4,6)$ is not on the curve. The point of the curve at $x = 4$ has $y = \pm\sqrt{32} \approx \pm 5.66$.

Here $x$ is a quadratic in $y$, so the parabola opens sideways. Complete the square in $y$, factoring the $2$ out first.

$x = 2(y^2 + 4y) + 5 = 2\bigl((y+2)^2 - 4\bigr) + 5 = 2(y+2)^2 - 3$

Isolating the square gives the standard form:

$(y + 2)^2 = \tfrac12 (x + 3)$

Comparing with $(y-k)^2 = 4p(x-h)$, the vertex is $(h, k) = (-3, -2)$. Write the coordinates in the usual order, $x$ first, even though $y$ is the squared variable.

Work with $x^2 = 4py$. The focal width is the chord through the focus parallel to the directrix, of length $|4p|$, while the focus $(0,p)$ and the directrix $y = -p$ are separated by

$|p - (-p)| = |2p|.$

Divide:

$\frac{|4p|}{|2p|} = 2$

The ratio is $2$ for every parabola, whatever its size or direction. It is a shape fact: all parabolas are similar to one another, so any two lengths defined this way keep a fixed ratio.