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Parabolas as a Distance Locus: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two perpendicular distances

Case A: Find the perpendicular foot and distance from $P$ at $(8,-2)$ to $y=4$ .

Case B: Find the perpendicular foot and distance from $Q$ at $(-5,-7)$ to $x=-1$ .
1. Hint 1
  Identify whether each directrix is horizontal or vertical.
2. Hint 2
  Keep the coordinate parallel to the line and change the other to reach it.
Answer

Case A: foot $(8,4)$ , distance $6$ . Case B: foot $(-1,-7)$ , distance $4$ .

Full solution

Case A: The perpendicular is vertical, so the foot is $(8,4)$ .
The vertical gap has length
$|{-2}-4|=6$

Case B: The perpendicular is horizontal, so the foot is $(-1,-7)$ .
The horizontal gap has length
$|{-5}-(-1)|=4$

Answer

Case A: foot $(8,4)$ , distance $6$ . Case B: foot $(-1,-7)$ , distance $4$ .

Key idea

Distance to a coordinate-parallel line uses one perpendicular coordinate gap.
Problem 2 Locate the middle of the spacing

A parabola has focus $(5,6)$ and directrix $y=-2$ . Find its vertex, axis, and vertex-focus distance.

The focus and directrix before finding the vertex. The text supplies exact coordinates; d names the directrix.

Text description of this figure

Equal horizontal and vertical scales. Focus F at (5, 6). Directrix y = -2. The focus and directrix are shown without a parabola curve.
1. Hint 1
  The axis passes through the focus and is perpendicular to the directrix.
2. Hint 2
  The vertex is halfway from the focus to its foot on the directrix.
Answer

Vertex $(5,2)$ ; axis $x=5$ ; vertex-focus distance $4$ .

Full solution

The foot from $(5,6)$ to the directrix is $(5,-2)$ .
Their midpoint has height
$(6+(-2))/2=2$

The vertex is $(5,2)$ and the vertical axis is $x=5$ .
The focus is four units above the vertex.

The vertex is halfway along the axis. The text supplies exact coordinates; d names the directrix.

Answer

Vertex $(5,2)$ ; axis $x=5$ ; vertex-focus distance $4$ .

Key idea

The vertex bisects the perpendicular spacing between the focus and directrix.
Problem 3 Check two proposed locus points

A parabola has focus $F$ at $(-2,1)$ and directrix $x=4$ . Decide whether each point is on the locus. Give both distances in each case.

Point A: $(0,1)$ .

Point B: $(-2,7)$ .
1. Hint 1
  A point is on the locus exactly when its focus and directrix distances agree.
2. Hint 2
  The line distance is horizontal. For the focus distance, check whether one coordinate already agrees.
Answer

A: focus distance $2$ , line distance $4$ , off the locus. B: both distances $6$ , on the locus.

Full solution

Point A: Its height agrees with the focus, giving focus distance
$|0-(-2)|=2$

Point A: Its perpendicular distance to the directrix is
$|0-4|=4$

Point A: The distances differ, so A is off the locus.

Point B: Its x-coordinate agrees with the focus, giving focus distance
$|7-1|=6$

Point B: Its perpendicular distance to the directrix is
$|-2-4|=6$

Point B: They agree, so B is on the locus.

B has equal focus and perpendicular line distances. The text supplies exact coordinates; d names the directrix.

Answer

A: focus distance $2$ , line distance $4$ , off the locus. B: both distances $6$ , on the locus.

Key idea

Membership depends on the two actual distances, not the visual closeness of a point to a curve.
Problem 4 Write the locus from its geometry

The focus of a parabola is $(6,-3)$ and its directrix is $x=2$ . Find its equation, showing the distance equality from which it follows.

The given focus is right of the directrix. The text supplies exact coordinates; d names the directrix.

Text description of this figure

Equal horizontal and vertical scales. Focus F at (6, -3). Directrix x = 2. The focus and directrix are shown without a parabola curve.
1. Hint 1
  A possible point must have the same distance from the focus as from the line.
2. Hint 2
  Write the focus distance using both coordinate gaps and the line distance using its perpendicular gap.
3. Hint 3
  Square the equality and cancel matching terms.
Answer

$(y+3)^2=8(x-4)$ ; equivalent equation forms are accepted.

Full solution

For a possible point $(x,y)$ , the distance equality is
$\sqrt{(x-6)^2+(y+3)^2}=|x-2|$

Both sides are nonnegative.
Squaring and canceling the x-squared terms gives
$\begin{aligned}(x-6)^2+(y+3)^2&=(x-2)^2,\\(y+3)^2&=8x-32,\\(y+3)^2&=8(x-4).\end{aligned}$

The vertex is $(4,-3)$ and the signed parameter is $2$ .
Reversing the squared equality and taking nonnegative square roots recovers the defining distances.

Answer

$(y+3)^2=8(x-4)$ ; equivalent equation forms are accepted.

Key idea

An equation is the distance condition written for every possible point.
Problem 5 Sketch from the defining geometry

For $(x+6)^2=-4(y-1)$ , find the vertex, focus, and directrix.

Sketch the curve using the vertex and its two points at height $y=0$ . Give those two points.
1. Hint 1
  The x-square indicates a vertical axis. Read the vertex and signed parameter.
2. Hint 2
  At the requested height, the squared gap determines two horizontal positions.
Answer

Vertex $(-6,1)$ ; focus $(-6,0)$ ; directrix $y=2$ ; points $(-8,0)$ and $(-4,0)$ ; opens down.

Full solution

The vertex is $(-6,1)$ .
The signed parameter is $-1$ , so the focus is $(-6,0)$ and directrix is $y=2$ .

At $y=0$ , the equation becomes
$(x+6)^2=4$

The gaps from $x=-6$ are $-2$ and $2$ .
The points are $(-8,0)$ and $(-4,0)$ .
Each is 2 units from the focus and 2 units from the directrix.

Plot these points and the vertex, then draw a smooth downward curve symmetric about $x=-6$ .

The vertex and two checked points guide the downward sketch. The text supplies exact coordinates; d names the directrix.

Answer

Vertex $(-6,1)$ ; focus $(-6,0)$ ; directrix $y=2$ ; points $(-8,0)$ and $(-4,0)$ ; opens down.

Key idea

Exact equidistant points support a sketch without treating the drawing as proof.
Problem 6 Translate the whole defining configuration

A parabola has equation $(y-4)^2=20(x+1)$ . Translate it $3$ units left and $6$ units down. Give the new equation, focus, and directrix.
1. Hint 1
  A translation moves the vertex, focus, and directrix together.
2. Hint 2
  The signed vertex-to-focus step stays the same.
3. Hint 3
  Apply the horizontal move to the directrix as well as the vertex.
Answer

Focus $(1,-2)$ ; equation
$(y+2)^2=20(x+4)$
Directrix
$x=-9$

Full solution

The original vertex is $(-1,4)$ and the signed parameter is $5$ .
The original focus is $(4,4)$ and directrix is $x=-6$ .

The vertex moves to $(-4,-2)$ , while the focus moves to $(1,-2)$ .
Moving the vertical directrix left three units gives $x=-9$ .

The parameter remains $5$ , so the translated equation is
$(y+2)^2=20(x+4)$

The new focus is 5 units right of the new vertex and the directrix is 5 units left, checking the translated geometry.

Answer

Focus $(1,-2)$ ; equation
$(y+2)^2=20(x+4)$
Directrix
$x=-9$

Key idea

A translation preserves the defining distances while changing the location.
Problem 7 Recover geometry from an expanded equation

A parabola has equation $x^2+2x+16y+49=0$ . Find its vertex, axis, focus, directrix, and opening direction.
1. Hint 1
  A completed square identifies the vertex before you locate the focus.
2. Hint 2
  Complete the x-square and factor the remaining linear expression.
3. Hint 3
  The multiplier of the unsquared gap is four times the signed parameter.
Answer

Vertex $(-1,-3)$ ; focus $(-1,-7)$ ; opens down. Axis
$x=-1$
Directrix
$y=1$

Full solution

Complete the x-square by adding 1 on both sides.
$\begin{aligned}x^2+2x&=-16y-49,\\(x+1)^2&=-16y-48,\\(x+1)^2&=-16(y+3).\end{aligned}$

The vertex is $(-1,-3)$ and the axis is $x=-1$ .
The multiplier is $-16$ , so the signed parameter is $-4$ .

The focus is four units below the vertex at $(-1,-7)$ .
The directrix is four units above the vertex, so its equation is $y=1$ .
The curve opens down.

The recovered vertical geometry has the focus below the vertex. The text supplies exact coordinates; d names the directrix.

Answer

Vertex $(-1,-3)$ ; focus $(-1,-7)$ ; opens down. Axis
$x=-1$
Directrix
$y=1$

Key idea

Completing the square exposes the coordinate gaps that locate the parabola.
Problem 8 Recover focus data from one point

A parabola has directrix $y=-3$ and focus $(2,f)$ , with the focus off the directrix. It passes through $(6,2)$ . Find every possible $f$ and the corresponding parabola equation.
1. Hint 1
  The known point has a fixed distance from the directrix.
2. Hint 2
  Its horizontal gap from the focus is 4. Use that and the fixed distance to find possible vertical gaps.
3. Hint 3
  For each focus, locate the vertex before writing the equation.
Answer

Focus height $-1$ :
$(x-2)^2=4(y+2)$
Focus height $5$ :
$(x-2)^2=16(y-1)$

Full solution

The line distance from $(6,2)$ is $|2+3|=5$ .
Its squared focus distance must therefore be 25.
$16+(2-f)^2=25$

Thus $(2-f)^2=9$ , giving focus heights $-1$ or $5$ .
Both differ from $-3$ , so both positions are allowed.

For $f=-1$ , the vertex is $(2,-2)$ and the signed parameter is 1.
The equation is
$(x-2)^2=4(y+2)$

For $f=5$ , the vertex is $(2,1)$ and the signed parameter is 4.
The equation is
$(x-2)^2=16(y-1)$

At $(6,2)$ , the first equation gives 16 on both sides, as does the second.
In either configuration the focus distance is the hypotenuse of a right triangle with legs 4 and 3, so it is 5.

Answer

Focus height $-1$ :
$(x-2)^2=4(y+2)$
Focus height $5$ :
$(x-2)^2=16(y-1)$

Key idea

One point and a directrix can leave two possible focus positions.
Problem 9 Use spacing and direction together

A parabola has vertex $(3,-4)$ , a horizontal axis, and focus-directrix spacing $7$ . Its focus is left of its vertex. Find the equation, focus, and directrix.
1. Hint 1
  The vertex bisects the focus-to-directrix spacing along the axis.
2. Hint 2
  The direction from vertex to focus determines the sign of the parameter.
Answer

Focus $(-1/2,-4)$ ; equation
$(y+4)^2=-14(x-3)$
Directrix
$x=13/2$

Full solution

The vertex-focus distance is half of 7, or $7/2$ .
The focus is left, so $p=-7/2$ .
The coefficient $4p$ is $-14$ .

The horizontal-axis equation is
$(y+4)^2=-14(x-3)$

The focus x-coordinate is $-1/2$ .
The directrix x-coordinate is $13/2$ .
Their difference is 7, checking the given spacing.

Half of the full spacing lies on each side of the vertex. The text supplies exact coordinates; d names the directrix.

Answer

Focus $(-1/2,-4)$ ; equation
$(y+4)^2=-14(x-3)$
Directrix
$x=13/2$

Key idea

Spacing determines the size of the parameter while opening direction determines its sign.
Problem 10 Link the equation back to the distances

A parabola has equation $(y+1)^2=c(x-2)$ with nonzero $c$ .

Its directrix is $x=5$ . Find $c$ , the vertex, and the focus.

Check whether $P$ at $(-1,5)$ belongs to this parabola using both its equation and the two defining distances.
1. Hint 1
  Read the vertex first, then use the directrix to recover the signed parameter.
2. Hint 2
  The directrix is three units right of the vertex, so the focus is equally far to its left.
3. Hint 3
  Evaluate the equation and the perpendicular line distance separately.
Answer

$c=-12$ ; vertex $(2,-1)$ ; focus $(-1,-1)$ . $P$ is on the locus; both distances are $6$ and both equation sides are $36$ .

Full solution

The coordinate gaps give vertex $(2,-1)$ .
The directrix is three units right of the vertex, so the focus is three units left.
The signed parameter is $-3$ .

The coefficient is four times the signed parameter.
$c=-12$
The focus is $(-1,-1)$ .

For $P$ at $(-1,5)$ , the left side of the equation is
$(5+1)^2=36$

The right side is
$-12(-1-2)=36$
Thus P satisfies the equation.

The focus and P share x-coordinate $-1$ , so their distance is
$|5-(-1)|=6$

The perpendicular foot is $(5,5)$ and the line distance is
$|-1-5|=6$

The equal distances verify that the recovered focus and directrix define the same locus as the recovered equation.

The recovered focus and directrix give the same six-unit distance. The text supplies exact coordinates; d names the directrix.

Answer

$c=-12$ ; vertex $(2,-1)$ ; focus $(-1,-1)$ . $P$ is on the locus; both distances are $6$ and both equation sides are $36$ .

Key idea

Recover the geometry and then check a point in both representations.

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