Precalculus · G.GPE.A.2

The parabola as a locus

A satellite dish catches a faint signal over its whole face and brings all of it to one small receiver. What shape does that, and where exactly does the receiver go?

You need

Focus and directrixPrecalculus Vertex form: reading, converting and graphingAlgebra 1 The distance formulaMath 8

Unlocks

Identifying and translating conicsPrecalculus

Where you are 6 in Conic sections

Geometry11

Conic sections6

Conic sections from a double cone Focus and directrix The ellipse The hyperbola

The parabola as a locus

Identifying and translating conics

Why this exists

Vertex form describes a parabola as the graph of y = a(x - h)^2 + k: a vertex, a stretch, and a curve that opens up or down. That says what the curve looks like, but not what makes it special, and it cannot describe a parabola that opens sideways.

Focus and directrix gave the deeper description: the conic with eccentricity e = 1 is every point as far from a fixed point as from a fixed line. This lesson works out that case in full: an equation for each of the four directions, the focus read from the equation, and why dishes, solar cookers and flashlight reflectors are all cut in this one shape.

Definition

Fix a point F, the focus, and a line L that does not pass through F, the directrix. The parabola with this focus and directrix is the set of all points P in the plane with

PF = PL,

where PF is the distance from P to F (the distance formula) and PL is the distance from P to the line L, measured along the perpendicular.

The parts of a parabola:

The axis is the line through F perpendicular to L. The parabola is symmetric about it.
The vertex is the point of the parabola on the axis. It is halfway between F and L.
The number p is the distance from the vertex to the focus, given a sign: positive when the focus lies in the positive direction (+y or +x) from the vertex, negative otherwise. The directrix is the same distance |p| on the other side of the vertex.
The focal width (or latus rectum) is the chord through the focus parallel to the directrix. Its length is |4p|.

Intuition

Figure 1Every point P of the parabola is as far from the focus F as from the directrix, PF = PD.

The vertex is the one point exactly halfway between focus and directrix. Move sideways from it and the distance to the focus grows, so the point must also move away from the directrix to keep the two distances equal: the curve bends away from the directrix and wraps around the focus. It never closes.

The distance |p| sets the size. A focus close to its directrix gives a narrow parabola, a focus far away a wide one. All parabolas are the same shape at different sizes, as all circles are.

Results

Standard equations, vertex at the origin. With p \neq 0:

x^2 = 4py has focus (0, p) and directrix y = -p. It opens up if p > 0, down if p < 0.
y^2 = 4px has focus (p, 0) and directrix x = -p. It opens right if p > 0, left if p < 0.

The squared variable names the axis it runs across: x^2 means a vertical axis, y^2 a horizontal one. The parabola always opens toward the focus.

Vertex at (h, k).

(x - h)^2 = 4p(y - k) has focus (h, k + p) and directrix y = k - p.
(y - k)^2 = 4p(x - h) has focus (h + p, k) and directrix x = h - p.

Link to vertex form. y = a(x - h)^2 + k is the same as \displaystyle (x - h)^2 = \frac{1}{a}(y - k), so \displaystyle 4p = \frac{1}{a} and \displaystyle p = \frac{1}{4a}. The focus is \displaystyle \left(h, k + \frac{1}{4a}\right).

Focal width. The chord through the focus has length |4p|; for x^2 = 4py its ends are (\pm 2p, p).

Reflective property. A ray traveling parallel to the axis that strikes the inside of the parabola reflects through the focus. Run backward, a ray leaving the focus reflects into a ray parallel to the axis.

Why it is true

The equation. Put the vertex at the origin, the focus at F = (0, p) and the directrix at y = -p. For a point P = (x, y),

PF = \sqrt{x^2 + (y - p)^2}, \qquad PL = |y + p|.

Both sides of PF = PL are at least 0, so squaring loses nothing and adds nothing:

\begin{aligned} x^2 + y^2 - 2py + p^2 &= y^2 + 2py + p^2 \\ x^2 &= 4py. \end{aligned}

Exchanging the roles of x and y (focus (p, 0), directrix x = -p) gives y^2 = 4px. Sliding the whole picture h right and k up changes no distance, so the vertex-(h, k) forms follow by writing x - h for x and y - k for y.

Focal width. On x^2 = 4py put y = p: x^2 = 4p^2, so x = \pm 2p, and the chord has length |4p|.

The reflective property. Take P on the parabola and let D be the foot of the perpendicular from P to the directrix, so PF = PD. Let t be the perpendicular bisector of FD. Since PF = PD, P lies on t. Any other point Q of t has QF = QD, and QD is longer than the perpendicular distance QL because QD is not perpendicular to L (only the segment through P is). So QF > QL: every point of t except P lies outside the parabola, on the side away from the focus. The line t meets the parabola at P only and otherwise stays on one side of it. For a curve bending one way, that is what tangent means here. Meeting once is not enough (the line through P parallel to the axis meets the curve once but crosses it). Calculus later defines tangents in general, and it picks out this same line.

In the isosceles triangle FPD, the perpendicular bisector of the base FD also bisects the angle at P, so t makes equal angles with PF and with PD. A ray parallel to the axis arrives along the line DP, from the side opposite D. The angle it makes with t is vertical to the angle between PD and t, so it equals the angle between PF and t. Light reflects with equal angles to the surface, so the reflected ray runs along PF, into the focus.

Worked examples

Example 1

The equation from the definition

Find the equation of the parabola with focus (0, 5) and directrix y = -5. Find the ends of its focal width and check one of them against the definition.

Solution

For P = (x, y), the distance formula gives \displaystyle PF = \sqrt{x^2 + (y - 5)^2}. The directrix is horizontal, so the perpendicular from P is vertical and PL = |y + 5|. Set them equal and square (both sides are at least 0):

\begin{aligned} x^2 + y^2 - 10y + 25 &= y^2 + 10y + 25 \\ x^2 &= 20y. \end{aligned}

This is x^2 = 4py with p = 5, as it must be: the vertex (0, 0) is halfway between the focus and the directrix. The focal width runs along y = 5: x^2 = 100, so its ends are (\pm 10, 5) and it has length 20 = 4p.

Check (10, 5): its distance to (0, 5) is 10, and its distance to y = -5 is 5 - (-5) = 10. Equal.

Example 2

A parabola that opens sideways

Find the vertex, focus, directrix and focal width of y^2 = -16x, and sketch it.

Solution

The squared variable is y, so this is y^2 = 4px and the axis is horizontal (the x-axis). Matching coefficients, 4p = -16, so p = -4. Since p < 0 the parabola opens left. The vertex is (0, 0), the focus (p, 0) = (-4, 0), and the directrix x = -p, that is x = 4.

The focal width is |4p| = 16. At x = -4, y^2 = 64, so its ends are (-4, \pm 8).

Figure 2Example 2. y^2 = -16x opens left, toward its focus and away from its directrix.

Check a point: (-1, 4) satisfies 16 = -16(-1). Its distance to the focus is \displaystyle \sqrt{3^2 + 4^2} = 5, and to the directrix x = 4 it is |-1 - 4| = 5.

Example 3

A coefficient to divide out

Find the focus and directrix of 2y^2 = 5x.

Solution

Divide both sides by 2 to reach the standard form: \displaystyle y^2 = \frac{5}{2}x. Now \displaystyle 4p = \frac{5}{2}, so \displaystyle p = \frac{5}{8}. The axis is the x-axis and p > 0, so it opens right, with focus \displaystyle \left(\frac{5}{8}, 0\right) and directrix \displaystyle x = -\frac{5}{8}.

The focal width is \displaystyle \frac{5}{2}, with ends \displaystyle \left(\frac{5}{8}, \pm\frac{5}{4}\right). Check: \displaystyle \left(\frac{5}{4}\right)^2 = \frac{25}{16} and \displaystyle \frac{5}{2} \cdot \frac{5}{8} = \frac{25}{16}.

Example 4

Focus and directrix away from the origin

Find the equation of the parabola with focus (-3, 4) and directrix x = 1.

Solution

The directrix is vertical, so the axis is the horizontal line through the focus, y = 4, and the equation has the form (y - k)^2 = 4p(x - h). The vertex is halfway along the axis between the focus (-3, 4) and the directrix point (1, 4): the midpoint is (-1, 4). The focus lies 2 units left of the vertex, so p = -2 and the parabola opens left:

(y - 4)^2 = -8(x + 1).

The definition gives the same equation directly. Squaring \displaystyle \sqrt{(x + 3)^2 + (y - 4)^2} = |x - 1|,

\begin{aligned} x^2 + 6x + 9 + (y - 4)^2 &= x^2 - 2x + 1 \\ (y - 4)^2 &= -8x - 8 = -8(x + 1). \end{aligned}

Example 5

Completing the square first

Find the vertex, focus and directrix of x^2 + 4x - 6y + 22 = 0.

Solution

Only x is squared, so the axis is vertical. Keep the x terms on the left and move the rest: x^2 + 4x = 6y - 22. Add 4 to both sides to complete the square:

(x + 2)^2 = 6y - 18 = 6(y - 3).

So h = -2, k = 3 and 4p = 6, giving \displaystyle p = \frac{3}{2}. The parabola opens up from the vertex (-2, 3), the focus is \displaystyle \left(-2, \frac{9}{2}\right), and the directrix is \displaystyle y = \frac{3}{2}.

In vertex form the same curve is \displaystyle y = \frac{1}{6}(x + 2)^2 + 3, with \displaystyle a = \frac{1}{6}, and \displaystyle p = \frac{1}{4a} = \frac{6}{4} = \frac{3}{2} agrees.

Example 6

Where the receiver goes

Uchechi’s satellite dish is 60 cm across at its rim and 9 cm deep. Its cross section through the center is a parabola. How far from the bottom of the dish must the receiver sit?

Solution

The signal arrives in rays parallel to the axis, and the reflective property gathers them at the focus, so the receiver belongs there. Put the vertex (the bottom of the dish) at the origin with the axis pointing up, so the cross section is x^2 = 4py. The rim is 30 cm either side of the center at height 9 cm, so (30, 9) is on the curve:

30^2 = 4p(9), \qquad 900 = 36p, \qquad p = 25.

The cross section is x^2 = 100y, and the receiver sits 25 cm above the bottom of the dish, which is 25 - 9 = 16 cm beyond the plane of the rim.

Figure 3Example 6. Rays parallel to the axis all reflect to the focus, 25 cm above the bottom of the dish.

Example 7

A tangent line and a reflected ray

Snorri aims a ray straight down the line x = 20 onto the inside of x^2 = 20y. Find the tangent line at the point it strikes, and show the ray reflects through the focus.

Solution

The ray strikes P = (20, 20), since 20^2 = 400 = 20 \cdot 20. From Example 1 the focus is F = (0, 5) and the directrix y = -5, so D = (20, -5).

The tangent is the perpendicular bisector of FD. The midpoint of FD is (10, 0), and FD has slope \displaystyle \frac{-5 - 5}{20 - 0} = -\frac{1}{2}, so the tangent has slope 2 and is y = 2(x - 10), that is y = 2x - 20. As a check, substituting into x^2 = 20y gives x^2 - 40x + 400 = 0, that is (x - 20)^2 = 0: the line meets the parabola only at x = 20.

Figure 4Example 7. The tangent at P makes equal angles with the incoming ray and with PF, so the ray reflects to the focus.

Also \displaystyle PF = \sqrt{20^2 + 15^2} = 25 and PD = 20 - (-5) = 25, so triangle FPD is isosceles and the tangent bisects the angle at P. The incoming ray meets the tangent at the same angle as PD, hence as PF, so it reflects along PF into the focus.

Boundaries

The focus must not lie on the directrix. If F is on L, then PF \geq PL for every P, with equality only on the line through F perpendicular to L. The set collapses to that line, which is why every standard form needs p \neq 0.

Only axes parallel to a coordinate axis. The four standard forms assume a horizontal or vertical directrix. A slanted directrix gives a parabola whose equation has an xy term, which these forms cannot show.

A sideways parabola is not a function of x. y^2 = 4px fails the vertical line test: each x on the open side gives two values \displaystyle y = \pm\sqrt{4px}.

The reflective property is for rays parallel to the axis. A ray arriving at a slant reflects somewhere other than the focus, which is why a dish must point straight at its signal source.

Where this goes next

Identifying conics reads an equation like x^2 + 4x - 6y + 22 = 0 at a glance: one squared variable means a parabola. The ellipse and the hyperbola have reflective properties of their own, between two foci.

Exercises

Reading a parabola with vertex at the origin

Find the direction of opening, focus, directrix and focal width of x^2 = 28y.
Find the direction of opening, focus, directrix and focal width of y^2 = 10x.
Find the direction of opening, focus, directrix and focal width of x^2 = -2y.
Find the direction of opening, focus, directrix and focal width of 2y^2 + 9x = 0.

Writing the equation, vertex at the origin

Write the equation of the parabola with focus (0, -6) and directrix y = 6.
Write the equation of the parabola with focus (9, 0) and directrix x = -9.
A parabola has its vertex at the origin, opens right, and passes through (2, 6). Find its equation and its focus.
A parabola has its vertex at the origin, a vertical axis, and passes through (-6, -4). Find its equation, focus and directrix.

Vertex away from the origin

Find the vertex, focus and directrix of (x - 1)^2 = 16(y + 2).
Find the vertex, focus and directrix of (y + 3)^2 = -10(x - 4).
Write the equation of the parabola with focus (2, 5) and directrix y = 1.
Write the equation of the parabola with focus (-1, -2) and directrix x = 3.
Find the vertex, focus and directrix of y^2 + 4y - 4x + 16 = 0.
Find the vertex, focus and directrix of x^2 - 10x + 2y + 21 = 0.
Find the focus and directrix of y = 2(x - 3)^2 + 1.

Dishes and reflectors

Tryggvi’s solar cooker is a parabolic dish 120 cm across and 20 cm deep. How far from the bottom of the dish should the pot sit?
Snorri’s flashlight reflector is a parabola with the bulb at its focus, 2.5 cm from the vertex. The reflector is 10 cm deep. How wide is its opening?
Uchechi has a second dish, 72 cm across, with its receiver 18 cm from the bottom. How deep is the dish, and where is the receiver compared with the rim?
Find the tangent line to x^2 = 20y at (30, 45) as the perpendicular bisector of FD, and check that it meets the parabola only once.

Reasoning

Explain why the vertex is the point of a parabola closest to its focus.
Explain what the set PF = PL becomes when the focus lies on the directrix.
Show that the focal width of y^2 = 4px has length |4p|.
Show that every graph y = ax^2 + bx + c with a \neq 0 is a parabola with a focus and directrix, and give p in terms of a.
Uchechi says that a dish of the same width but twice the depth has its receiver twice as far from the bottom. Test her claim on the dish of Example 6 made 18 cm deep, and explain the result.

Answers

Opens up, focus (0, 7), directrix y = -7, focal width 28. Here 4p = 28, so p = 7.
Opens right, focus \displaystyle \left(\frac{5}{2}, 0\right), directrix \displaystyle x = -\frac{5}{2}, focal width 10. Here 4p = 10, so \displaystyle p = \frac{5}{2}.
Opens down, focus \displaystyle \left(0, -\frac{1}{2}\right), directrix \displaystyle y = \frac{1}{2}, focal width 2. Here 4p = -2, so \displaystyle p = -\frac{1}{2}.
Opens left, focus \displaystyle \left(-\frac{9}{8}, 0\right), directrix \displaystyle x = \frac{9}{8}, focal width \displaystyle \frac{9}{2}. Rearranged, \displaystyle y^2 = -\frac{9}{2}x, so \displaystyle 4p = -\frac{9}{2} and \displaystyle p = -\frac{9}{8}.
x^2 = -24y. The vertex is halfway, at the origin, and the focus is 6 below it, so p = -6.
y^2 = 36x. The vertex is the origin and the focus is 9 to its right, so p = 9.
y^2 = 18x, focus \displaystyle \left(\frac{9}{2}, 0\right). From 6^2 = 4p(2), 36 = 8p and \displaystyle p = \frac{9}{2}.
x^2 = -9y, focus \displaystyle \left(0, -\frac{9}{4}\right), directrix \displaystyle y = \frac{9}{4}. From (-6)^2 = 4p(-4), 36 = -16p and \displaystyle p = -\frac{9}{4}.
Vertex (1, -2), focus (1, 2), directrix y = -6. Here 4p = 16, so p = 4 and it opens up.
Vertex (4, -3), focus \displaystyle \left(\frac{3}{2}, -3\right), directrix \displaystyle x = \frac{13}{2}. Here 4p = -10, so \displaystyle p = -\frac{5}{2} and it opens left.
(x - 2)^2 = 8(y - 3). The vertex is midway between (2, 5) and (2, 1), at (2, 3), and the focus is 2 above it, so p = 2.
(y + 2)^2 = -8(x - 1). The vertex is midway between (-1, -2) and (3, -2), at (1, -2), and the focus is 2 to its left, so p = -2.
Vertex (3, -2), focus (4, -2), directrix x = 2. From y^2 + 4y = 4x - 16, adding 4 gives (y + 2)^2 = 4(x - 3), so p = 1.
Vertex (5, 2), focus \displaystyle \left(5, \frac{3}{2}\right), directrix \displaystyle y = \frac{5}{2}. From x^2 - 10x = -2y - 21, adding 25 gives (x - 5)^2 = -2(y - 2), so \displaystyle p = -\frac{1}{2}.
Focus \displaystyle \left(3, \frac{9}{8}\right), directrix \displaystyle y = \frac{7}{8}. With a = 2, \displaystyle p = \frac{1}{4a} = \frac{1}{8} from the vertex (3, 1).
45 cm. The rim point is (60, 20) on x^2 = 4py, so 3600 = 80p and p = 45.
20 cm. With p = 2.5 the reflector is x^2 = 10y; at depth y = 10, x^2 = 100 and x = \pm 10.
18 cm deep, so the receiver is exactly level with the rim. The dish is x^2 = 72y, and at x = 36, \displaystyle y = \frac{1296}{72} = 18.
y = 3x - 45. With F = (0, 5) and D = (30, -5), the midpoint is (15, 0) and FD has slope \displaystyle -\frac{1}{3}, so the tangent has slope 3. Substituting, x^2 - 60x + 900 = 0, which is (x - 30)^2 = 0: one point only.
Since PF = PL, the point nearest F is the point nearest L. For x^2 = 4py with p > 0, every point has y \geq 0, so PL = y + p \geq p, with equality only at the vertex. So PF is smallest, |p|, at the vertex. The other directions work the same way.
A line. For any P, PF \geq PL, since PL is the shortest distance from P to the line and F is a point of it. Equality holds only when the perpendicular from P lands at F, which is the line through F perpendicular to L.
The chord through the focus (p, 0) runs along x = p. Then y^2 = 4p^2, so y = \pm 2p, and the chord’s ends (p, 2p) and (p, -2p) are |4p| apart.
Completing the square writes y = ax^2 + bx + c as y = a(x - h)^2 + k, which is \displaystyle (x - h)^2 = \frac{1}{a}(y - k). That is the standard form with \displaystyle 4p = \frac{1}{a}, so \displaystyle p = \frac{1}{4a}: focus \displaystyle \left(h, k + \frac{1}{4a}\right), directrix \displaystyle y = k - \frac{1}{4a}.
Her claim fails: the receiver moves closer, to \displaystyle \frac{25}{2} = 12.5 cm. From 30^2 = 4p(18), \displaystyle p = \frac{900}{72} = \frac{25}{2}. In general \displaystyle p = \frac{r^2}{4d} for rim radius r and depth d, so doubling the depth halves p: a deeper dish curves faster and focuses closer.