Unit 8, lesson 92

Parabolas as Geometric Loci

How a parabola is defined as the set of points equally far from a focus and a directrix, and how to find its vertex, focus, directrix and latus rectum from an equation.

  • Theory part 183
  • Practice part 184
  • Conic Sections

Shine a flashlight straight at a wall and the edge of the light is a circle. Tilt the flashlight until one edge of its cone of light runs parallel to the wall, and the edge on the wall becomes a curve that opens up and never closes. That curve is a parabola. It also has a description that needs no cone at all, only one point and one line, and that description is what turns it into an equation.

Key terms

Conic section

Definition Let a line, the generator, cross a fixed line, the axis, at a point VV, the vertex of the cone, at an angle α\alpha with 0<α<π20 < \alpha < \frac{\pi}{2}. Turning the generator about the axis sweeps out a right circular double cone. A conic section is the curve in which a plane that does not pass through VV meets this cone. If β\beta is the angle between the plane and the axis, with 0≤β≤π20 \le \beta \le \frac{\pi}{2}, the curve is

Angle of the planeConic section
β=π2\beta = \frac{\pi}{2}circle
α<β<π2\alpha < \beta < \frac{\pi}{2}ellipse that is not a circle
β=α\beta = \alpha, parallel to a generatorparabola
0≤β<α0 \le \beta < \alphahyperbola, with two branches

A plane through VV gives a degenerate section instead, which is a single point, a single line or a pair of crossing lines.

In plain words Slice a double cone with a flat plane. A level cut gives a circle, a tilted cut gives an ellipse, a cut parallel to the side of the cone gives a parabola, and a steep cut through both halves gives a hyperbola.

Example The cone x2+y2=z2x^2 + y^2 = z^2 has the zz-axis as its axis and α=π4\alpha = \frac{\pi}{4}. Here are three slices.

PlaneAfter substitutingCurve
z=3z = 3x2+y2=9x^2 + y^2 = 9circle
z=x+2z = x + 2y2=4(x+1)y^2 = 4(x + 1)parabola
x=1x = 1z2−y2=1z^2 - y^2 = 1hyperbola

The plane z=x+2z = x + 2 makes the angle π4\frac{\pi}{4} with the zz-axis, the same angle as the cone, and it is parallel to the generator z=xz = x in the plane y=0y = 0. Substituting z=x+2z = x + 2 into x2+y2=z2x^2 + y^2 = z^2 cancels x2x^2 and leaves y2=4x+4y^2 = 4x + 4. So the shadow of this curve on the xyxy-plane is a parabola. The curve itself, which lies in the tilted plane, is a parabola too, because tilting the shadow back up only stretches it by the factor 2\sqrt{2} along its axis.

Parabola

Definition Let FF be a point and LL a line in the plane, with FF not on LL. The parabola with focus FF and directrix LL is the set of all points PP in the plane whose distance to FF equals their distance to LL,

∣PF∣=dist⁡(P,L),\lvert PF \rvert = \operatorname{dist}(P, L),

where ∣PF∣\lvert PF \rvert is the distance d(P,F)d(P, F) between the two points and dist⁡(P,L)\operatorname{dist}(P, L) is the perpendicular distance from PP to LL. Ellipses that are not circles and hyperbolas have a similar description, in which the distance to the focus is a fixed multiple ee of the distance to the directrix, with 0<e<10 < e < 1 for an ellipse and e>1e > 1 for a hyperbola. That multiple is the eccentricity, and the parabola is the case e=1e = 1. This description returns with the polar equations of conics. The parabolas cut from a cone and the parabolas defined this way are the same curves.

In plain words Stand anywhere on a parabola. The straight line distance to the focus equals the shortest distance to the directrix.

Example Take F=(0,2)F = (0, 2) and let LL be the line y=−2y = -2. A point at height 22 is 44 units above LL, so it must also be 44 units from FF. Since FF is at the same height, that gives (4,2)(4, 2) and (−4,2)(-4, 2). A point at height 88 is 1010 units above LL, and

x2+(8−2)2=10\sqrt{x^2 + (8 - 2)^2} = 10

gives x2=64x^2 = 64, so (8,8)(8, 8) and (−8,8)(-8, 8) are on the parabola. Collecting such points one height at a time traces out the curve.

Directrix

Definition The directrix of a parabola is the fixed line LL in its definition. Distances to LL are measured perpendicular to LL. For a horizontal line y=cy = c the distance from (x,y)(x, y) is ∣y−c∣\lvert y - c \rvert, and for a vertical line x=cx = c it is ∣x−c∣\lvert x - c \rvert. The parabola never meets its directrix and lies entirely on the same side of it as the focus. A point on LL has distance 00 to LL but a positive distance to FF. For a point PP on the other side, the segment from PP to FF crosses LL at a point QQ, so dist⁡(P,L)≤∣PQ∣<∣PF∣\operatorname{dist}(P, L) \le \lvert PQ \rvert < \lvert PF \rvert.

In plain words The directrix is a line the parabola bends away from. Every point of the curve is on the focus side of the line, and the curve opens away from it.

Example Take the focus (0,−2)(0, -2) and the directrix y=2y = 2. The point (4,−2)(4, -2) is 44 units from the focus and ∣−2−2∣=4\lvert -2 - 2 \rvert = 4 units from the directrix, so it is on the parabola. The point (0,3)(0, 3), on the other side of the directrix, is only 11 unit from the line but 55 units from the focus, so it is not. This parabola opens downward, away from its directrix.

Focus

Definition The focus of a parabola is the fixed point FF in its definition. It lies on the parabola’s axis of symmetry, the line through the focus perpendicular to the directrix, inside the curve, on the side toward which the parabola opens. Because every point PP of the parabola has ∣PF∣=dist⁡(P,L)\lvert PF \rvert = \operatorname{dist}(P, L), the distance from PP to the focus, called its focal distance, can be read from the distance to the directrix without a square root.

In plain words The focus is the point the parabola wraps around. To find how far a point of the curve is from it, measure the distance to the directrix instead, which is much easier.

Example Take F=(0,1)F = (0, 1) and the directrix y=−1y = -1. The point (6,9)(6, 9) is 1010 units above the directrix, and its distance to the focus is

(6−0)2+(9−1)2=36+64=10,\begin{aligned} \sqrt{(6 - 0)^2 + (9 - 1)^2} &= \sqrt{36 + 64} \\[0.4em] &= 10, \end{aligned}

so it is on the parabola. Every point (x,y)(x, y) of this parabola lies above the directrix, so its focal distance is simply y+1y + 1, with no square root and no absolute value.

Vertex and axis

Definition The axis of a parabola is the line through the focus perpendicular to the directrix, and the parabola is symmetric about it. The vertex is the point where the parabola meets its axis. It is the midpoint between the focus and the point where the axis meets the directrix. It is the point of the parabola closest to the focus and also the closest to the directrix, at the same distance from both. The distance from the vertex to the focus is written ∣p∣\lvert p \rvert with p≠0p \ne 0, and the sign of pp records which way the parabola opens, as in the standard equations below.

In plain words The axis is the mirror line of the parabola, and the vertex is its tip, exactly halfway between the focus and the directrix.

Example Take the focus (2,5)(2, 5) and the directrix y=1y = 1. The axis is the vertical line x=2x = 2, and it meets the directrix at (2,1)(2, 1). The vertex is the midpoint of (2,5)(2, 5) and (2,1)(2, 1), which is (2,3)(2, 3). It is 22 units from the focus and 22 units from the directrix, and the focus is above it, so p=2p = 2.

Standard equations

Definition Let p≠0p \ne 0. The parabola with focus (0,p)(0, p) and directrix y=−py = -p has vertex (0,0)(0, 0) and equation

x2=4py.x^2 = 4py.

It opens upward if p>0p > 0 and downward if p<0p < 0. In the same way, the parabola with focus (p,0)(p, 0) and directrix x=−px = -p is y2=4pxy^2 = 4px, which opens right if p>0p > 0 and left if p<0p < 0. For the first one, the definition says

x2+(y−p)2=∣y+p∣.\sqrt{x^2 + (y - p)^2} = \lvert y + p \rvert.

Both sides are nonnegative, so squaring gives the equivalent equation

x2+y2−2py+p2=y2+2py+p2,\begin{aligned} &x^2 + y^2 - 2py + p^2 \\[0.4em] &\quad = y^2 + 2py + p^2, \end{aligned}

which simplifies to x2=4pyx^2 = 4py. Moving the vertex to (h,k)(h, k) gives

(x−h)2=4p(y−k),(y−k)2=4p(x−h).\begin{gathered} (x - h)^2 = 4p(y - k), \\[0.4em] (y - k)^2 = 4p(x - h). \end{gathered}

The first has focus (h,k+p)(h, k + p) and directrix y=k−py = k - p. The second has focus (h+p,k)(h + p, k) and directrix x=h−px = h - p.

In plain words The number pp is the signed distance from the vertex to the focus. Its sign tells you which way the parabola opens, and the directrix sits the same distance away on the other side of the vertex.

Example The parabola with focus (2,5)(2, 5) and directrix y=1y = 1 has vertex (2,3)(2, 3) and p=2p = 2, so its equation is

(x−2)2=8(y−3).(x - 2)^2 = 8(y - 3).

The point (6,5)(6, 5) satisfies it, since 16=8⋅216 = 8 \cdot 2. It is also 44 units from the focus and 44 units from the directrix, as it should be. The parabola with focus (0,2)(0, 2) and directrix y=−2y = -2 from earlier is x2=8yx^2 = 8y, and the points (±4,2)(\pm 4, 2) and (±8,8)(\pm 8, 8) found there satisfy it.

Latus rectum

Definition The latus rectum of a parabola is the chord through the focus perpendicular to the axis. For x2=4pyx^2 = 4py it lies on the line y=py = p, its endpoints are (−2p,p)(-2p, p) and (2p,p)(2p, p), and its length is

4∣p∣.4\lvert p \rvert.

For y2=4pxy^2 = 4px the endpoints are (p,−2p)(p, -2p) and (p,2p)(p, 2p). Each endpoint is 2∣p∣2\lvert p \rvert from the focus and also 2∣p∣2\lvert p \rvert from the directrix.

In plain words The latus rectum measures how wide the parabola is level with the focus. Together with the vertex, its two endpoints give three points for a quick and accurate sketch.

Example For y2=10xy^2 = 10x we have 4p=104p = 10, so p=52p = \frac{5}{2}. The focus is (52,0)(\frac{5}{2}, 0) and the directrix is x=−52x = -\frac{5}{2}. The latus rectum runs from (52,−5)(\frac{5}{2}, -5) to (52,5)(\frac{5}{2}, 5) and has length 1010. As a check, 52=25=10⋅525^2 = 25 = 10 \cdot \frac{5}{2}, and the endpoint (52,5)(\frac{5}{2}, 5) is 55 units from the focus and 55 units from the directrix.

Check yourself

  1. Find the focus, the directrix and the length of the latus rectum of x2=−16yx^2 = -16y. Which way does it open?
Show the answer

Comparing with x2=4pyx^2 = 4py gives 4p=−164p = -16, so p=−4p = -4. The focus is (0,−4)(0, -4), the directrix is y=4y = 4, the latus rectum has length 1616, and the parabola opens downward.

  1. Write the equation of the parabola with focus (−1,4)(-1, 4) and directrix x=3x = 3.
Show the answer

The axis is the horizontal line y=4y = 4, which meets the directrix at (3,4)(3, 4). The vertex is the midpoint (1,4)(1, 4), and the focus is 22 units to its left, so p=−2p = -2. The equation is

(y−4)2=−8(x−1),(y - 4)^2 = -8(x - 1),

and the parabola opens to the left.

  1. A point on x2=20yx^2 = 20y is 1313 units from the focus. Find the point.
Show the answer

Here 4p=204p = 20, so p=5p = 5 and the directrix is y=−5y = -5. The focal distance equals the distance to the directrix, y+5=13y + 5 = 13, so y=8y = 8. Then x2=160x^2 = 160 and x=±410x = \pm 4\sqrt{10}. The two points are (410,8)(4\sqrt{10}, 8) and (−410,8)(-4\sqrt{10}, 8).

How this lesson connects

Where this leads in Calculus