The focus-directrix definition of a parabola, the two standard equations with vertex at the origin, identifying the focus, directrix and axis of symmetry from an equation, writing an equation from a given focus or directrix, and applying the model to parabolic reflectors.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Graph and Write Equations of Parabolas
Objectives
Five outcomes. A parabola redefined by distance, which opens up two directions it could not go before.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 620-623 — the lesson these objectives are drawn from
Warm-up
Chapter 4 defined a parabola as the graph of a quadratic function, opening up or down.
Discussion prompt
Every quadratic function opens up or down. Is there any way for the graph of a function to open left or right?
Hint: Think about the vertical line test.
Answer:
No. A curve opening left or right fails the vertical line test — one input would give two outputs — so it cannot be a function at all.
\[ y^2 = 4px \;\Longrightarrow\; y = \pm\sqrt{4px}, \text{ two values} \]
So this lesson needs a definition that does not mention functions. The focus-directrix definition is that definition, and it makes all four directions equally natural.
Concept
A parabola is the set of all points equally far from a fixed point, the focus, and a fixed line, the directrix. The vertex lies halfway between them, and each is the same distance from the vertex.
focus and directrix — The fixed point and fixed line that define a parabola. Every point of the curve is the same distance from the focus as from the directrix, and both lie the same distance from the vertex.
\[ x^2 = 4py: \; \text{focus } (0,p), \; \text{directrix } y = -p \]
Defining the curve by distances rather than by a formula is what lets a parabola open in any direction, and it is how every conic section in this chapter will be defined.
Figure (svg): A parabola with its focus and directrix, and two points shown equally far from each
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 620-621
Section
Section 1
Concept
Each point on a parabola is the same distance from the focus as it is from the directrix. The focus lies on the axis of symmetry, the directrix is perpendicular to it, and the vertex sits halfway between.
\[ \text{dist to focus} = \text{dist to directrix} \]
The distance to a line means the perpendicular distance, which for a horizontal directrix is just the vertical gap. That is why the checks in this lesson are so quick.
Figure (svg): A parabola with its focus and directrix, and two points shown equally far from each
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 620-620 — Focus and directrix
Picture it
A parabola with two of its points and their distances marked.
Figure (svg): A parabola with its focus and directrix, and two points shown equally far from each
The nearer point is 2 from each; the farther one is 5 from each. Every point on the curve behaves the same way, which is what the definition says.
Worked example
Verifying the definition for the parabola whose equation is x squared equals 4y.
\[ \text{For } x^2 = 4y, \; \text{focus } (0,1), \; \text{directrix } y=-1, \text{ check the points } (2,1) \text{ and } (4,4). \]
First point: distance to the focus
Why: From 2 comma 1 to 0 comma 1 is 2 across.
\[ 2 \]
First point: distance to the directrix
Why: From height 1 down to height negative 1.
\[ 2 \]
Second point: distance to the focus
Why: Four across and 3 up, so 16 plus 9.
\[ \sqrt{25} = 5 \]
Second point: distance to the directrix
Why: From height 4 down to height negative 1.
\[ 5 \]
Figure (svg): A parabola with its focus and directrix, and two points shown equally far from each
\[ 2 = 2 \quad \text{and} \quad 5 = 5 \]
Verify: check the vertex too
Why: The vertex at the origin is 1 unit from the focus and 1 unit from the directrix, so it satisfies the definition as well — and it is the point where the two distances are smallest. That is why the vertex sits exactly halfway between the focus and the directrix, at distance the absolute value of p from each.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 620-620
Fill the middle
The point where x is 4 and y is 4, with directrix y equal to negative 1.
Fill in the blanks
\text5 = 4 - (-1) = ___
Why: The perpendicular distance to a horizontal line is the difference of the heights, which is 5. That matches the distance to the focus, as the definition requires.
Worked example
Reading a parabola's features from the definition.
\[ \text{For } y^2 = 4px \text{ with } p>0, \text{ name the vertex, focus, directrix and axis of symmetry.} \]
The vertex
Why: The standard forms of this lesson all have it at the origin.
\[ (0, 0) \]
The axis of symmetry
Why: Y is squared, so the curve is symmetric about the horizontal axis.
\[ y = 0 \]
The focus
Why: On the axis of symmetry, p units from the vertex.
\[ (p, 0) \]
The directrix
Why: Perpendicular to the axis, p units the other way.
\[ x = -p \]
Figure (svg): The solution to Worked example locate the parts on a graph shown as a ladder of expressions, one row per algebraic move
\[ (0,0), \; y = 0, \; (p,0), \; x = -p \]
Verify: check the vertex is halfway
Why: The focus is at x equal to p and the directrix at x equal to negative p, so the midpoint of the horizontal gap between them is x equal to 0 — the vertex. That halfway relationship holds in all four cases and is the fastest way to recover p from a picture.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621
Trap
\[ \text{point } (4,4), \; \text{directrix } y = -1 \]
Measure to a convenient point on the directrix
Why: The distance is taken to where the directrix meets the axis of symmetry.
\[ \sqrt{16+25} = \sqrt{41} \quad \text{(wrong)} \]
The distance from a point to a LINE is the perpendicular distance, the shortest one — not the distance to some particular point on it.
\[ \text{perpendicular distance} = 4 - (-1) = 5 \]
Measure straight down to a horizontal directrix
Why: Perpendicular to a horizontal line means vertical.
\[ 5 = 5 \quad \checkmark, \text{ matching the distance to the focus} \]
For a vertical directrix the perpendicular distance is horizontal instead. Either way it is a single subtraction, not a distance formula.
Matching
Four parts of every parabola.
Match the pairs
Why: The focus is always inside the curve and the directrix always outside it, on opposite sides of the vertex. That opposition is why the directrix equation carries a minus sign that the focus does not.
Prediction
Commit before reasoning.
Predict first
Why does the vertex lie exactly midway between the focus and the directrix?
Correct: Because the vertex is on the curve, so its two distances are equal, and it sits on the axis.
\[ \text{focus at } p, \; \text{directrix at } -p \;\Longrightarrow\; \text{vertex at } 0 \]
Why: Being on the parabola forces its distance to the focus to equal its distance to the directrix; being on the axis of symmetry means both of those distances are measured along that axis. A point on a line, equidistant from another point on that line and from a perpendicular line, must be at the midpoint. So the halfway property is a consequence of the definition rather than an extra rule, and it holds in all four orientations.
Sorting
The focus and the directrix sit on opposite sides.
Sort into buckets
Sort each item by where it lies relative to a parabola opening upward.
Knowing which side each lies on catches a sign error immediately: a directrix that cuts through the curve has been placed on the wrong side.
Section
Section 2
Concept
With vertex at the origin, a parabola has equation x squared equals 4py when it opens up or down, and y squared equals 4px when it opens left or right. The sign of p decides which of the two directions.
\[ x^2 = 4py \quad \text{or} \quad y^2 = 4px \]
The four cases are one form seen twice, with the roles of x and y exchanged. Parabolas opening left or right are not functions, which is why Chapter 4 could not reach them.
Figure (svg): The four standard parabolas with vertex at the origin, two vertical and two horizontal
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621 — Standard Equation of a Parabola with Vertex at the Origin
Picture it
The standard parabolas with vertex at the origin.
Figure (svg): The four standard parabolas with vertex at the origin, two vertical and two horizontal
Reading left to right: up, down, right, left. The squared variable picks the axis and the sign of p picks the direction along it.
Worked example
The key concept's table, applied.
\[ \text{For } x^2 = 4py \text{ and } y^2 = 4px, \text{ give the focus, directrix and axis in each case.} \]
Vertical form: the axis
Why: X is squared, so the curve is symmetric about the vertical axis.
\[ x = 0 \]
Vertical form: focus and directrix
Why: P units up and p units down from the vertex.
\[ (0, p)\text{ and } y = -p \]
Horizontal form: the axis
Why: Y is squared, so the symmetry is about the horizontal axis.
\[ y = 0 \]
Horizontal form: focus and directrix
Why: P units right and p units left.
\[ (p, 0)\text{ and } x = -p \]
Figure (svg): The four standard parabolas with vertex at the origin, two vertical and two horizontal
\[ (0,p), y=-p, x=0; \quad (p,0), x=-p, y=0 \]
Verify: check the sign convention
Why: In both forms the directrix carries the opposite sign to the focus, since they lie on opposite sides of the vertex. A negative p simply swaps which side each is on, so the formulas need no separate case for it — one pair of rules covers all four pictures.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621
Sorting
Squared variable, then sign of p.
Sort into buckets
Sort each equation by the direction it opens.
Two independent readings, each taking a second: which variable carries the square, and what sign sits on the other side.
Worked example
Guided Practice 1 to 4.
\[ \text{Put } y^2=-6x, \; x^2=2y, \; y=-\tfrac{1}{4}x^2, \; x=\tfrac{1}{3}y^2 \text{ into standard form and find } p. \]
First: already standard
Why: Four p equals negative 6.
\[ p = -\frac{3}{2},\text{ opens left} \]
Second: already standard
Why: Four p equals 2.
\[ p = \frac{1}{2},\text{ opens up} \]
Third: multiply by negative 4
Why: X squared equals negative 4y, so 4p is negative 4.
\[ p = -1,\text{ opens down} \]
Fourth: multiply by 3
Why: Y squared equals 3x, so 4p is 3.
\[ p = \frac{3}{4},\text{ opens right} \]
Figure (svg): The solution to Worked example convert four equations to standard form shown as a ladder of expressions, one row per algebraic move
\[ p = -\tfrac{3}{2}, \; \tfrac{1}{2}, \; -1, \; \tfrac{3}{4} \]
Verify: check one direction against the equation
Why: For the third, y equals negative a quarter x squared is a downward parabola from Chapter 4, and p came out negative — consistent. Whenever the equation is solvable for y, the Chapter 4 reading and the Chapter 9 reading must agree, and comparing them catches a sign slip at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622
Error analysis
A student identifies p from a parabola in standard form.
Annotate
On: \( x^2 = 2y \;\Longrightarrow\; p = 2, \; \text{focus } (0,2) \)
The 4 is there so that p measures the distance from the vertex to the focus directly. Dropping it multiplies every distance by four.
Fill the middle
Guided Practice 2.
Fill in the blanks
x^2 = 2y: \; 4p = 2 \;\Longrightarrow\; p = \frac2___}
Why: Two divided by 4 is one half, so p is one half and the focus is half a unit above the vertex. Forgetting to divide by 4 is the commonest slip in this idea.
Comparison
Fill the blanks. Swap x and y throughout.
Comparison matrix
| Feature | x^2 = 4py | y^2 = 4px |
|---|---|---|
| Opens | up or down | left or right |
| Focus | (0, p) | (p, 0) |
| Directrix | y = -p | x = -p |
| Is it a function? | yes | no |
The last row is what Chapter 4 could not accommodate. Defining the curve by distance rather than by a rule for outputs is what removes the restriction.
Prediction
Commit before reasoning.
Predict first
Why is the standard form written x squared equals 4py rather than x squared equals py?
Correct: So that p measures the distance from the vertex to the focus directly.
\[ x^2 = 4py: \; |p| = \text{ distance from vertex to focus} \]
Why: With the 4 in place, the focus is at exactly p units and the directrix at exactly p units the other way, so a single letter reads off both. Without it every distance would need dividing by 4 at each use. The same convention will appear in the circle, ellipse and hyperbola forms of the coming lessons: the constants are arranged so that each letter measures something you can point at on the graph.
Section
Section 3
Concept
Rewrite the equation in standard form, read off p, then state the focus, the directrix and the axis of symmetry. Plot points on the side the parabola opens toward.
\[ x = -\tfrac{1}{8}y^2 \;\Longrightarrow\; y^2 = -8x, \; p = -2 \]
Which inputs are usable follows from the sign of p. With p negative in the horizontal form, only x values at most zero produce real y values.
Figure (svg): A leftward-opening parabola with its focus, directrix and axis of symmetry marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621 — Graph an equation of a parabola
Picture it
Example 1, with all three features drawn.
Figure (svg): A leftward-opening parabola with its focus, directrix and axis of symmetry marked
The focus at negative 2 comma 0 sits inside the curve and the directrix at x equal to 2 outside it, two units from the vertex on each side.
Worked example
Example 1.
\[ \text{Graph } x = -\tfrac{1}{8}y^2 \text{ and identify the focus, directrix and axis of symmetry.} \]
Rewrite in standard form
Why: Multiply both sides by negative 8.
\[ y ^{2} = -8 x \]
Find p
Why: Four p equals negative 8.
\[ p = -2 \]
State the features
Why: Y is squared, so the axis is horizontal.
\[ (-2, 0), x = 2, y = 0 \]
Plot points
Why: P is negative, so only non-positive x gives real y.
\[ x = -1\text{ gives } y\text{ about plus or minus } 2.83 \]
Figure (svg): A leftward-opening parabola with its focus, directrix and axis of symmetry marked
\[ (-2,0), \; x = 2, \; y = 0 \]
Verify: check the definition at one plotted point
Why: Take x equal to negative 2, where y is plus or minus 4. From negative 2 comma 4 to the focus at negative 2 comma 0 is 4 straight down. To the directrix at x equal to 2 is 4 across. Equal, as the definition demands — and this point is the endpoint of the chord through the focus, which is always 4 times the absolute value of p long.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621
Fill the middle
Example 1.
Fill in the blanks
x = -\tfrac-8___y^2 \;\Longrightarrow\; y^2 = ___x
Why: Multiplying by negative 8 isolates y squared and gives 4p equal to negative 8, so p is negative 2. Getting to standard form is always the first move.
Worked example
Guided Practice 1 to 4.
\[ \text{Identify the focus, directrix and axis for } y^2=-6x, \; x^2=2y, \; y=-\tfrac{1}{4}x^2, \; x=\tfrac{1}{3}y^2. \]
First: p is negative three halves
Why: Horizontal form, opening left.
\[ (-\frac{3}{2}, 0), x = \frac{3}{2}, y = 0 \]
Second: p is one half
Why: Vertical form, opening up.
\[ (0, \frac{1}{2}), y = -\frac{1}{2}, x = 0 \]
Third: rewrite as x squared equals negative 4y
Why: P is negative 1, opening down.
\[ (0, -1), y = 1, x = 0 \]
Fourth: rewrite as y squared equals 3x
Why: P is three quarters, opening right.
\[ (\frac{3}{4}, 0), x = -\frac{3}{4}, y = 0 \]
Figure (svg): The solution to Worked example four more sets of features shown as a ladder of expressions, one row per algebraic move
\[ (-\tfrac{3}{2},0); \; (0,\tfrac{1}{2}); \; (0,-1); \; (\tfrac{3}{4},0) \]
Verify: check that each directrix is on the far side
Why: In every case the directrix lies on the opposite side of the vertex from the focus, and at the same distance. For the first, the focus is 1.5 to the left and the directrix 1.5 to the right. A directrix that came out on the same side as the focus would mean a lost minus sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622
Trap
\[ y^2 = -8x, \; \text{try } x = 2 \]
Make a table of positive x values
Why: The habit from Chapter 4, where x ran both ways, is carried over.
\[ y^2 = -16 \;\Longrightarrow\; \text{no real } y \quad \text{(no points at all)} \]
With p negative the parabola opens LEFT, so every point on it has x at most zero. Positive inputs give nothing.
\[ \text{use } x = -1, -2, -3, -4, -5 \]
Choose inputs on the side the curve opens toward
Why: The sign of p says which half of the axis carries the graph.
\[ x = -2 \;\Longrightarrow\; y^2 = 16 \;\Longrightarrow\; y = \pm 4 \]
Each usable input gives TWO points, above and below the axis, which is exactly why these are not functions.
Matching
Divide the coefficient by 4.
Match the pairs
Why: Two of the four foci sit on the horizontal axis and two on the vertical, matching which variable was squared. In every case the coordinate is the coefficient divided by 4.
Ranking
Graphing a parabola from its equation.
Put in order
Why: Step four depends on step two: the sign of p is what tells you which inputs are usable, and choosing the wrong half of the axis produces a table with no real values in it at all.
Prediction
Commit before reasoning.
Predict first
For y squared equals negative 8x, how far apart are the two points on the curve directly above and below the focus?
Correct: 8 units, which is 4 times the absolute value of p.
\[ x = p \;\Longrightarrow\; y^2 = 4p^2 \;\Longrightarrow\; y = \pm 2|p| \]
Why: At x equal to negative 2, the focus's x-coordinate, the equation gives y squared equal to 16, so y is plus or minus 4 — a gap of 8. In general substituting x equal to p gives y squared equal to 4p squared, so the two points are 2 times the absolute value of p above and below, a total of 4 times it. That chord is called the latus rectum, and it gives a quick second pair of points for any sketch.
Section
Section 4
Concept
With the vertex at the origin, a focus or a directrix determines p, and its position determines which standard form to use. Substituting p into that form finishes the job.
\[ \text{directrix } y = -\tfrac{3}{2} \;\Longrightarrow\; p = \tfrac{3}{2} \;\Longrightarrow\; x^2 = 6y \]
A focus on the vertical axis or a horizontal directrix means the vertical form; a focus on the horizontal axis or a vertical directrix means the horizontal one.
Figure (svg): A parabola given by its vertex and directrix, with the equation recovered
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621 — Write an equation of a parabola
Picture it
Example 2: vertex at the origin and directrix below it.
Figure (svg): A parabola given by its vertex and directrix, with the equation recovered
The directrix is negative p, so p is three halves. Substituting into the vertical form gives x squared equals 6y.
Worked example
Example 2.
\[ \text{Write the equation of the parabola with vertex } (0,0) \text{ and directrix } y = -\tfrac{3}{2}. \]
Choose the form
Why: The directrix is horizontal, so the axis is vertical.
\[ x ^{2} = 4 p y \]
Find p
Why: Negative p equals negative three halves.
\[ p = \frac{3}{2} \]
Substitute
Why: Four times three halves is 6.
\[ x ^{2} = 6 y \]
Sanity-check the direction
Why: P is positive, so the curve opens up, away from the directrix.
Figure (svg): A parabola given by its vertex and directrix, with the equation recovered
\[ x^2 = 6y \]
Verify: check a point against the definition
Why: At x equal to 3 the equation gives y equal to 1.5. That point is 3 across and 0 up from the focus at 0 comma 1.5, a distance of 3; and it is 1.5 plus 1.5, or 3, above the directrix. Equal, so the equation really does describe the right curve.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 621-621
Fill the middle
Example 2.
Fill in the blanks
-p = -\tfrac3/2___ \;\Longrightarrow\; p = ___
Why: The directrix is y equal to negative p, so negative p equals negative three halves and p is positive three halves. The sign change is the whole step, and skipping it flips the curve.
Worked example
Guided Practice 5 to 8.
\[ \text{Write equations with vertex } (0,0) \text{ and: directrix } y=2; \; x=4; \; \text{focus } (-2,0); \; (0,3). \]
First: horizontal directrix above
Why: Negative p equals 2, so p is negative 2.
\[ x ^{2} = -8 y \]
Second: vertical directrix
Why: Negative p equals 4, so p is negative 4.
\[ y ^{2} = -16 x \]
Third: focus on the horizontal axis
Why: P is negative 2, so use the horizontal form.
\[ y ^{2} = -8 x \]
Fourth: focus on the vertical axis
Why: P is 3, so use the vertical form.
\[ x ^{2} = 12 y \]
Figure (svg): The solution to Worked example four equations from clues shown as a ladder of expressions, one row per algebraic move
\[ x^2=-8y, \; y^2=-16x, \; y^2=-8x, \; x^2=12y \]
Verify: check the direction in each case
Why: A directrix above the vertex means the curve opens down, and the first answer has a negative coefficient — consistent. A focus to the left means the curve opens left, and the third is negative too. The curve always opens toward the focus and away from the directrix, which is a one-glance check on every sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622
Error analysis
A student writes an equation from a directrix.
Annotate
On: \( \text{directrix } y = -\tfrac{3}{2} \;\Longrightarrow\; p = -\tfrac{3}{2}, \; x^2 = -6y \)
The curve always opens away from its directrix. Checking that one fact catches this sign error every time.
Sorting
Look at whether the clue is horizontal or vertical.
Sort into buckets
Sort each clue.
The clue's orientation and the form's are always opposite for a directrix and always the same for a focus, which is worth noticing rather than rederiving.
Matching
Find p, then choose the form.
Match the pairs
Why: The two directrix clues both produced negative coefficients because both lines sat on the positive side of the origin, forcing the curve to open the other way. The focus clues keep the sign of the focus's coordinate.
Prediction
Commit before reasoning.
Predict first
A parabola has vertex at the origin and directrix y equal to 2. Which way does it open?
Correct: Down, away from the directrix and toward the focus.
\[ \text{directrix } y=2 \;\Longrightarrow\; p = -2 \;\Longrightarrow\; x^2 = -8y \]
Why: The focus and directrix sit on opposite sides of the vertex, so a directrix above means a focus below, and the curve always wraps around its focus. Algebraically, negative p equals 2 gives p equal to negative 2 and the equation x squared equals negative 8y, which is downward. The geometric reading and the algebraic one agree, and using them against each other is the best check available in this idea.
Section
Section 5
Concept
Energy arriving parallel to the axis of symmetry is reflected to the focus, and energy emitted from the focus leaves parallel to the axis. That property is why receivers and light sources are placed exactly at the focus.
\[ x^2 = 4py \text{ with } p = 4.5 \;\Longrightarrow\; x^2 = 18y \]
The focus-directrix definition is what makes it work: equal distances mean every path from a distant source to the focus has the same length, so the arriving waves stay in step.
Figure (svg): The cross section of a parabolic solar dish with its focus and depth marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622 — Parabolic reflectors
Picture it
Example 3: the EuroDish, with its engine at the focus.
Figure (svg): The cross section of a parabolic solar dish with its focus and depth marked
The focus is 4.5 metres above the vertex and the dish is 8.5 metres across, but only about 1 metre deep. A distant focus gives a shallow curve.
Worked example
Example 3.
\[ \text{A dish has its engine } 4.5 \text{ m above the vertex and is } 8.5 \text{ m wide. Find its equation and depth.} \]
Find p
Why: The focus is above the vertex, so p is positive.
\[ p = 4.5 \]
Write the equation
Why: Four times 4.5 is 18.
\[ x ^{2} = 18 y \]
Find the edge's x-value
Why: Half of 8.5 on each side of the vertex.
\[ x = 4.25 \]
Solve for the depth
Why: Four point two five squared is 18.0625, over 18.
\[ y\text{ about } 1.0 \]
Figure (svg): The cross section of a parabolic solar dish with its focus and depth marked
\[ x^2 = 18y, \quad \text{depth} \approx 1 \text{ m} \]
Verify: sanity-check the shape
Why: A dish 8.5 metres across and 1 metre deep is very shallow, which matches a focus far out at 4.5 metres. Halving the focal length to 2.25 would give x squared equals 9y and a depth of about 2 metres — twice as deep. Focal length and depth trade off directly, which is what makes this model useful for design.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622
Fill the middle
Example 3, Step 2.
Fill in the blanks
\text2 8.5 \text___ \;\Longrightarrow\; x = \frac______} = 4.25
Why: The axis of symmetry runs through the middle of the dish, so the edge is half the width from it. Using the full width would make the computed depth four times too big.
Worked example
The same model with a nearer focus.
\[ \text{A reflector } 8.5 \text{ m wide has its focus } 2 \text{ m above the vertex. Find its equation and depth.} \]
Find p and the equation
Why: Four times 2 is 8.
\[ x ^{2} = 8 y \]
Use the same half-width
Why: Four point two five metres from the vertex.
\[ x = 4.25 \]
Solve for the depth
Why: Eighteen point zero six two five over 8.
\[ y\text{ about } 2.26 \]
Compare
Why: Less than half the focal length gives more than twice the depth.
Figure (svg): The solution to Worked example a deeper dish shown as a ladder of expressions, one row per algebraic move
\[ x^2 = 8y, \quad \text{depth} \approx 2.26 \text{ m} \]
Verify: check the inverse relationship
Why: The depth at a fixed width is the half-width squared divided by 4p, so halving p a little more than doubles the depth — 4.5 gave 1.00 and 2 gives 2.26, a ratio of 2.25, which is exactly 4.5 over 2. The depth varies inversely with the focal length, which is Lesson 8.1's inverse variation in a new setting.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 622-622
Trap
\[ x^2 = 18y, \; \text{dish } 8.5 \text{ m wide} \]
Substitute the full width
Why: The stated width is used directly as the x-coordinate.
\[ (8.5)^2 = 18y \;\Longrightarrow\; y \approx 4.0 \quad \text{(wrong)} \]
The vertex is at the middle of the dish, so the edge is only half the width away — 4.25 metres, not 8.5.
\[ x = \tfrac{8.5}{2} = 4.25 \]
Use the half-width, since the vertex is centred
Why: The equation measures x from the axis of symmetry, which runs through the middle.
\[ (4.25)^2 = 18y \;\Longrightarrow\; y \approx 1.0 \]
Using the full width gives a depth four times too large, since the error is squared. Sketching the dish with the axis through its middle makes the halving obvious.
Prediction
Commit before reasoning.
Predict first
Why does a satellite dish put its receiver at the focus rather than at the vertex?
Correct: Because every parallel ray reflects to the focus, so all the arriving energy concentrates there.
\[ \text{parallel in} \;\Longrightarrow\; \text{all to } (0,p) \]
Why: A distant source sends rays that are effectively parallel to the axis, and the parabolic surface directs every one of them to the same point. Putting the receiver anywhere else collects only the small part of the energy that happens to pass through it. The equal-distance definition also means all those paths have the same total length, so the waves arrive in step rather than cancelling — which is why the shape must be a parabola and not merely a bowl.
Comparison
Fill the blanks. Same width, different focus.
Comparison matrix
| Quantity | Focus at 4.5 m | Focus at 2 m |
|---|---|---|
| Equation | x^2 = 18y | x^2 = 8y |
| Half-width | 4.25 m | 4.25 m |
| Depth | about 1.0 m | about 2.26 m |
| Shape | shallow | deeper, more curved |
Depth varies inversely with focal length at a fixed width, so a designer trades a compact deep dish against a shallow one with a distant, more awkwardly mounted receiver.
Ranking
From a physical dish to its depth.
Put in order
Why: Step one is a choice, and it is what makes p and the width mean what the formulas expect. Placing the vertex anywhere else would require the translated forms of Lesson 9.6 instead.
Comparison
Fill the blanks. Two readings settle every one.
Comparison matrix
| Equation | Opens | Focus |
|---|---|---|
| x^2 = 4py, p > 0 | up | (0, p) |
| x^2 = 4py, p < 0 | down | (0, p), below the vertex |
| y^2 = 4px, p > 0 | right | (p, 0) |
| y^2 = 4px, p < 0 | left | (p, 0), left of the vertex |
The squared variable picks the axis and the sign of p picks the direction along it. Nothing else varies across the four.
Pattern
One routine to graph, one to write.
The curve always opens toward the focus and away from the directrix. That single fact checks every sign in the lesson.
OpenStax Algebra and Trigonometry 2e, §12.3 The Parabola §12.3
Check
Standard form. Divide by 4.
Check your understanding
What is the focus of x^2 = 2y?
Answer: A
Why: 4p = 2 gives p = 1/2, and x is squared so the focus is on the vertical axis.
Check
Graphing. Which way does it open?
Check your understanding
What are the directrix and axis of symmetry of x = -(1/8)y^2?
Answer: A
Why: In standard form y^2 = -8x, so p = -2, and the directrix is x = -p.
Check
Writing an equation. Watch the sign.
Check your understanding
Write the equation with vertex (0, 0) and directrix y = 2.
Answer: A
Why: -p = 2 gives p = -2, and a horizontal directrix means the vertical form.
Real world
A car headlight has a parabolic reflector 12 centimetres across and 4 centimetres deep, with the bulb at the focus.
Discussion prompt
Find the equation of the cross section with the vertex at the origin, locate the bulb, and explain why the beam comes out parallel.
Hint: The rim is at x equal to 6 and y equal to 4.
Answer:
\[ x^2 = 4py, \; (6)^2 = 4p(4) \;\Longrightarrow\; 36 = 16p \;\Longrightarrow\; p = 2.25 \]
\[ x^2 = 9y, \quad \text{focus } (0, 2.25) \]
The bulb sits 2.25 centimetres from the back of the reflector, on the axis. Because the reflector is a parabola and the bulb is at its focus, every ray leaving the bulb bounces off parallel to the axis — the reverse of the solar dish, where parallel rays converged on the focus.
That is the whole reason headlights, torches and searchlights use this shape rather than a sphere or a cone: only a parabola turns a point source into a parallel beam. It also explains why a bulb that has shifted even a few millimetres from the focus produces a beam that spreads and dazzles — the geometry is unforgiving, which is why headlight alignment is a legal requirement.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is every parabola the graph of a function?
Correct: No — those opening left or right fail the vertical line test.
\[ y^2 = 4px \;\Longrightarrow\; y = \pm\sqrt{4px} \]
Why: Chapter 4 met only the parabolas y equals a x squared, which open up or down and are functions. But y squared equals 4px gives y equal to plus or minus the square root of 4px, two outputs for each usable input, so a vertical line crosses the curve twice. This is why the lesson needs the focus-directrix definition: it describes the curve by a distance property rather than by a rule assigning one output to each input, and that description works in all four directions equally.
Explain it
They know parabolas as the graphs of quadratics and have never heard of a focus.
Discussion prompt
In four sentences or fewer, explain what a focus and a directrix are.
Hint: Describe a rule for building the curve.
Answer:
Pick a point and a line that does not pass through it. Now mark every spot that is exactly as far from the point as it is from the line.
Those spots form a parabola, the point is called the focus and the line the directrix. The curve wraps around the focus and bends away from the directrix, and the vertex is the one spot exactly halfway between them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the 4, write 4p equals the coefficient as its own line before solving. For directrix signs, remember the curve opens away from the directrix and check your answer against that. For choosing the form, ask which variable is squared, or which way the axis of symmetry runs. For reflectors, sketch the dish with the axis through its middle so the halving of the width is visible.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a parabola page. Top: sketch all four standard parabolas in a row, labelling each with its equation, its focus, its directrix and its axis of symmetry, and write one sentence saying what the squared variable controls and what the sign of p controls. Middle left: draw the parabola x squared equals 4y, mark the focus and directrix, pick two points on the curve and verify the equal distances for both. Middle right: graph Example 1 in full, plotting at least four points and marking all three features. Bottom left: work Guided Practice 5 to 8, showing the sign change for each directrix clue. Bottom right: draw the EuroDish cross section, label the 8.5 metre width and the 4.5 metre focal length, and compute the depth, then repeat with a focus at 2 metres and note what changed.
If any of your four sketches has the directrix cutting through the curve, redo it: the directrix always lies entirely outside the parabola, on the far side of the vertex from the focus.
Recap
Five things, and a definition that will shape the rest of the chapter.
| If you see | Then |
|---|---|
| x squared on one side | Vertical axis of symmetry; use x^2 = 4py |
| y squared on one side | Horizontal axis; use y^2 = 4px |
| A coefficient c on the other side | 4p = c, so p = c/4 |
| A directrix | It is at -p, so change the sign |
| A focus | Its nonzero coordinate is p directly |
| A reflector's width | Halve it before substituting |
Lesson 9.3 does the same for circles, defining them by a single distance from a centre — which is the distance formula of Lesson 9.1 with the square roots cleared away.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.2 Graph and Write Equations of Parabolas §9.2, pp. 620-623 — everything on these slides traces back here
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