Parabolas as a Distance Locus

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Learning goals

Test whether a point is equally distant from a focus and directrix
Derive a parabola equation from its defining equality of distances
Write translated equations for vertical and horizontal parabola axes
Recover the focus, directrix, vertex, and opening direction from an equation
Verify recovered geometry using both distances and the equation

Equal distance to a point and a line

Distance to a line follows the perpendicular, shortest route. For $P$ at $(4,2)$ and the horizontal line $y=-2$ , that route ends at $H$ at $(4,-2)$ . Its length is $4$ . If the fixed point is $F$ at $(0,2)$ , the horizontal segment $PF$ also has length $4$ .

A locus is the set of all points satisfying a stated condition. A parabola is the locus of points equally distant from a fixed point, its focus, and a fixed line, its directrix. The focus must be off the directrix. We consider horizontal or vertical directrices here.

Equal distances trace an upward parabola. V, P, and Q each satisfy the same condition.

The figure traces three exact points. The middle point $V$ at $(0,0)$ is $2$ units from both $F$ and the directrix. Each of $P$ at $(4,2)$ and $Q$ at $(-4,2)$ is $4$ units from both. The perpendicular feet are $H$ at $(4,-2)$ and $I$ at $(-4,-2)$ . Each pair of distances is checked separately; all the segments do not have the same length.

For a horizontal line $y=c$ , the distance from $(x,y)$ is $|y-c|$ .

For a vertical line $x=c$ , it is $|x-c|$ . Absolute values keep these distances nonnegative, even below or left of the line.

Check your understanding

For $P$ at $(7,-1)$ and directrix $y=-5$ , what is the perpendicular foot?

(7,-5)

(-5,-1)

(7,5)

(0,-1)

Worked example 1 Test the distance condition

Problem. A parabola has focus $F$ at $(0,2)$ and directrix $y=-2$ . Decide whether each point lies on it.

Point $P$ at $(4,2)$ .

Point $B$ at $(2,1)$ .

Solution. Point $P$ is $4$ units from the directrix. Its focus distance is

PF=\sqrt{(4-0)^2+(2-2)^2}.

This is $4$ , so $P$ lies on the locus, as in the figure.

Point $B$ is $3$ units from the directrix. Its focus distance is

BF=\sqrt{(2-0)^2+(1-2)^2}.

This is $\sqrt5$ , which differs from $3$ . Point $B$ is off the locus. A point being near the drawn curve is not the test; the equality is.

An equation records the distance condition

The vertex lies halfway from the focus to its perpendicular foot on the directrix. The line through those two points is the axis of symmetry.

With focus $(0,3)$ and directrix $y=-3$ , the halfway point is $(0,0)$ . On the axis, a point’s distances balance only at that midpoint. Reflection across the axis preserves the focus and directrix and preserves both distances, so it preserves the locus. This explains both the vertex and the axis of symmetry in the familiar graph.

Worked example 2 Derive an equation and check the reverse

Problem. Find an equation for the locus with focus $(0,3)$ and directrix $y=-3$ . Explain why every point satisfying your equation meets the original distance condition.

Solution. For a possible point $(x,y)$ , write the two distances and equate them.

\sqrt{x^2+(y-3)^2}=|y+3|.

Both sides are nonnegative. Squaring and expanding gives

\begin{aligned} x^2+(y-3)^2&=(y+3)^2,\\ x^2+y^2-6y+9&=y^2+6y+9,\\ x^2&=12y. \end{aligned}

This is the equation. Solving it for $y$ gives the quadratic graph $y=x^2/12$ .

The distance condition produces the vertical quadratic graph. The text supplies exact coordinates; d names the directrix.

For the reverse check, start with $x^2=12y$ . Substitution gives

\begin{aligned} x^2+(y-3)^2&=12y+(y-3)^2,\\ x^2+(y-3)^2&=(y+3)^2. \end{aligned}

Taking nonnegative square roots returns the distance equality, including the absolute value on the line distance. Thus the equation describes exactly the locus, without extra points from squaring.

A translated vertical axis

The parameter measures a signed step from the vertex toward the focus. For vertex $(2,-1)$ and focus $(2,0)$ , that step is one unit and the directrix is $y=-2$ . A point’s squared focus distance is $(x-2)^2+y^2$ ; its squared line distance is $(y+2)^2$ . Their equality simplifies to

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

The coordinate gaps from the vertex have replaced the plain coordinates. This is how a translation enters the equation.

The translated vertical distance locus#

First take vertex $(-2,1)$ , focus $(-2,3)$ , and directrix $y=-1$ . The squared-distance equality becomes

\begin{aligned} (x+2)^2+(y-3)^2&=(y+1)^2,\\ (x+2)^2+y^2-6y+9&=y^2+2y+1,\\ (x+2)^2&=8(y-1). \end{aligned}

Conversely, replacing $(x+2)^2$ by $8(y-1)$ in the squared focus distance gives $(y+1)^2$ . Taking nonnegative square roots recovers the equality of distances.

The same steps work at any vertex $(h,k)$ . Let $p$ be a nonzero signed number. Take focus $(h,k+p)$ and directrix $y=k-p$ .

For a point $(x,y)$ , equidistance means

\sqrt{(x-h)^2+(y-k-p)^2}=|y-k+p|.

Squaring is reversible because these are nonnegative distances. Expanding the squares gives

\begin{aligned} (x-h)^2+(y-k-p)^2&=(y-k+p)^2,\\ (x-h)^2&=4p(y-k). \end{aligned}

The matching square and constant terms cancel; the two opposite cross terms give four times $p(y-k)$ . Nothing depended on the chosen numerical coordinates. Conversely, the last equation gives

(x-h)^2+(y-k-p)^2=(y-k+p)^2.

Nonnegative square roots give equal focus and perpendicular line distances. Both directions hold for either sign of $p$ .

For a vertical axis, the equation is therefore

\begin{gathered}(x-h)^2=4p(y-k),\\p\ne0.\end{gathered}

The vertex is $(h,k)$ and the axis is $x=h$ .

The focus is $(h,k+p)$ and the directrix is $y=k-p$ . The vertex-focus distance is $|p|$ , while the focus-directrix spacing is $2|p|$ .

For positive $p$ , the nonnegative square forces $y\ge k$ , so the curve opens up.

For negative $p$ , it forces $y\le k$ , so it opens down. The vertex is the curve’s point on its axis in either case. A negative $p$ indicates direction, not a negative distance. Zero is excluded because it would put the focus on the directrix.

Check your understanding

The locus has equation $x^2=20y$ . What is the vertex-focus distance?

20

10

5

4

Worked example 3 Use a focus below its directrix

Problem. The focus is $(3,-5)$ and the directrix is $y=1$ . Find the vertex, axis, signed parameter, equation, and opening direction.

Solution. The foot from the focus is $(3,1)$ . Their midpoint is $(3,-2)$ , the vertex. The axis is $x=3$ .

The focus is $3$ units below the vertex, so $p=-3$ . The equation is

(x-3)^2=-12(y+2).

It opens down. The distance from vertex to focus is $3$ , and the spacing from focus to directrix is $6$ .

Perpendicular feet lie above the traced points when the directrix is above the focus.

At $P$ at $(9,-5)$ and $Q$ at $(-3,-5)$ , the focus distances are $6$ and the perpendicular distances to $y=1$ are also $6$ . These exact points guide the sketch on both sides of the axis.

Check your understanding

Which is the focus of $(x+4)^2=-6(y-2)$ ?

(-4,7/2)

(-4,1/2)

(4,1/2)

(-4,-4)

A horizontal axis

A vertical directrix makes the perpendicular distance horizontal. For example, with focus $(-3,2)$ and directrix $x=1$ , the vertex is $(-1,2)$ . Here the signed step from vertex to focus is $-2$ along the x-direction.

Worked example 4 Turn the distance calculation sideways

Problem. Find the equation for focus $(-3,2)$ and directrix $x=1$ . Verify that $(-3,6)$ is on the locus.

Solution. The distance equality is

\sqrt{(x+3)^2+(y-2)^2}=|x-1|.

Square and cancel the matching x-squared terms.

\begin{aligned} (x+3)^2+(y-2)^2&=(x-1)^2,\\ x^2+6x+9+(y-2)^2&=x^2-2x+1,\\ (y-2)^2&=-8(x+1). \end{aligned}

The curve opens left from $(-1,2)$ . At $(-3,6)$ , the focus distance is $4$ and the perpendicular distance to $x=1$ is also $4$ .

The directrix is vertical, so the dashed distance segments are horizontal.

In general, the horizontal form is

\begin{gathered}(y-k)^2=4p(x-h),\\p\ne0.\end{gathered}

The vertex is $(h,k)$ and the axis is $y=k$ .

The focus is $(h+p,k)$ and the directrix is $x=h-p$ . To derive this form, equate squared distances as above:

\begin{aligned} (x-h-p)^2+(y-k)^2&=(x-h+p)^2,\\ (y-k)^2&=4p(x-h). \end{aligned}

Reversing these steps and taking nonnegative square roots proves the reverse direction, just as for the vertical form. For positive $p$ , $x\ge h$ and the curve opens right.

For negative $p$ , $x\le h$ and it opens left. In both orientations, the opening points from the vertex toward the focus.

Changing the sign of the signed step changes the horizontal opening direction.

Here the vertex is $(1,1)$ , the focus is $(3,1)$ , and the directrix is $x=-1$ .

The equation is

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

The points $(3,5)$ and $(3,-3)$ are each $4$ units from both the focus and the line. This right-opening case uses the same distance rule as the left-opening case.

Check your understanding

A parabola has focus $(4,-2)$ and directrix $x=-2$ . Which equation describes it?

(y+2)^2=24(x-1)

(x-1)^2=12(y+2)

(y+2)^2=-12(x-1)

(y+2)^2=12(x-1)

Recover and verify the geometry

Read the multiplier of the unsquared coordinate gap as $4p$ , then divide by four. If the equation is expanded, complete the square first.

Worked example 5 Recover a hidden focus and check a point

Problem. A parabola has equation

y^2-6y+12x+21=0.

Find its vertex, axis, focus, directrix, and opening direction. Verify $P$ at $(-4,9)$ using the defining distances.

Solution. Complete the y-square.

\begin{aligned} y^2-6y&=-12x-21,\\ (y-3)^2&=-12x-12,\\ (y-3)^2&=-12(x+1). \end{aligned}

The vertex is $(-1,3)$ and the axis is $y=3$ .

Since $4p=-12$ , the signed parameter is $-3$ .

The focus is $(-4,3)$ and the directrix is $x=2$ . The curve opens left.

Recovered geometry verifies the point using two distances. The text supplies exact coordinates; d names the directrix.

Point $P$ is $6$ units above the focus and $6$ units left of the directrix. Thus both defining distances are $6$ . Substitution into the completed equation also gives $36$ on each side.

Check your understanding

After completing the square in $y^2-2y+8x-15=0$ , what is the directrix?

x=0

y=3

x=4

y=-1

Parabolas as a Distance Locus

Learning goals

Equal distance to a point and a line

An equation records the distance condition

A translated vertical axis

A horizontal axis

Recover and verify the geometry

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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