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Parabola: focus and directrix

Read the idea, work independently, then explain what changed.

高二選擇性必修第一册(A版).pdf · 3.3 · PDF 135 / printed page 130

TOPIC 01

Parabola: focus and directrix

Derive a parabola from equal distances, track orientation under translation and use the definition in a distance optimization.

What you will be able to explain

Convert between y²=4ax and its focus/directrix.
Retain the absolute distance to a line when deriving a translated equation.
Check attainability in a focal-distance inequality.

Equal distances produce the quadratic relation

The defining distance often replaces a square root with a simple linear quantity on the curve.

The locus of points equidistant from (a,0) and x=−a satisfies y²=4ax when a≠0. The sign of a controls orientation.

\sqrt{(x-a)^2+y^2}=|x+a|\quad\Longrightarrow\quad y^2=4ax

Square two nonnegative distances.
$(x-a)^2+y^2=(x+a)^2$
Cancel x² and a², retaining the sign in 4ax.
$y^2=4ax$

When it applies

A nondegenerate parabola has a nonzero focal parameter.
A vertex is not a center of half-turn symmetry.

Counterexample: For y²=8x, (2,4) lies on the curve but its half-turn about the vertex, (−2,−4), does not.

Focus/directrix diagram ↔ equality of distances ↔ translated standard equation.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict which side contains the curve when a changes sign.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

The sampled point has equal distances to the focus and directrix. At a=0 the equation reduces to y²=0, a degenerate line.

Numerical data for the diagram; printable
Quantity	Value
focus / 焦點	1, 0
directrix x / 準線x坐標	-1
PF	2
distance to directrix / 到準線距離	2

Explain: The model computes focus distance and directrix distance at the same sampled point; a=0 is displayed as a degenerate line.

Transfer: Translate the vertex while keeping a fixed and explain which distances remain unchanged.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

Find the vertex, focus and directrix.

y^2=12x

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Compare with y²=4ax.

Hint 2

The focus and directrix are equally far from the vertex.

Worked solution

Read 4a=12.
$a=3$
Place the focus on the opening axis.
$V=(0,0),\ F=(3,0),\ d:x=-3$

Vertex (0,0), focus (3,0), directrix x=−3.

Checks and common pitfalls: The coefficient 12 is four times the focal distance.

Reasoning checklist · self / teacher assessment

Vertex (0,0), focus (3,0), directrix x=−3.
The coefficient 12 is four times the focal distance.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Derive the locus equidistant from focus (2,0) and directrix x=−2.

First step provided — complete the reasoning:Equate squared distances.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Distance to the line is |x+2|.

Hint 2

Both distances are nonnegative before squaring.

Worked solution

Equate squared distances.
$(x-2)^2+y^2=(x+2)^2$
Cancel matching terms and check x≥0 follows.
$y^2=8x$

y²=8x.

Checks and common pitfalls: Using signed x+2 without explaining nonnegativity can hide an invalid distance interpretation.

Reasoning checklist · self / teacher assessment

y²=8x.
Using signed x+2 without explaining nonnegativity can hide an invalid distance interpretation.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Find the point on y²=8x closest to the focus and its distance.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Focal distance equals distance to the directrix.

Hint 2

The curve forces x≥0.

Worked solution

The focus is (2,0) and the directrix x=−2.
$PF=x+2$
Minimize x on the parabola.
$x=0,\quad y=0,\ PF=2$

Distance 2 at the vertex (0,0).

Checks and common pitfalls: The focus is not on the parabola.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the focus y-coordinate.

x^2=20y

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The axis is vertical.

Hint 2

Read 4a=20.

Worked solution

Convert the coefficient to focal distance.
$a=20/4=5$
The focus lies in the opening direction.
$F=(0,5)$

Focus y-coordinate 5.

Checks and common pitfalls: Interchanging x and y also interchanges the focus coordinate.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Find the focus and directrix.

y^2=-8x

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The parameter a is negative.

Hint 2

Keep its sign when locating the focus.

Worked solution

Read a=−2.
$4a=-8$
The parabola opens left.
$F=(-2,0),\ d:x=2$

Focus (−2,0), directrix x=2.

Checks and common pitfalls: A negative opening coefficient is valid, not an empty locus.

Reasoning checklist · self / teacher assessment

Focus (−2,0), directrix x=2.
A negative opening coefficient is valid, not an empty locus.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

For a point (x,4) on y²=8x, find x.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute the y-coordinate.

Hint 2

Divide its square by 8.

Worked solution

Square y=4.
$16=8x$
Solve the linear equation.
$x=2$

x=2.

Checks and common pitfalls: Using y=4 directly instead of y² gives the wrong result.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Find the length of the chord through the focus perpendicular to the axis.

y^2=8x

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The focus is at x=2.

Hint 2

Find both y-coordinates.

Worked solution

Substitute x=2.
$y^2=16\Rightarrow y=\pm4$
Subtract the endpoint coordinates.
$L=4-(-4)=8$

Chord length 8.

Checks and common pitfalls: A single ordinate 4 is half the chord.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Find the vertex, focus, directrix and opening direction.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use shifted coordinates X=x+2,Y=y−1.

Hint 2

Translate focus and directrix back together.

Worked solution

In shifted coordinates a=2.
$V=(-2,1),\ F=V+(2,0)=(0,1)$
The directrix is two units left of the vertex.
$d:x=-4$

Vertex (−2,1), focus (0,1), directrix x=−4; opens right.

Checks and common pitfalls: The focus is not obtained by shifting only the x-coordinate of the equation coefficient.

Reasoning checklist · self / teacher assessment

Vertex (−2,1), focus (0,1), directrix x=−4; opens right.
The focus is not obtained by shifting only the x-coordinate of the equation coefficient.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Explain why the vertex of y²=8x is not a centre of symmetry.

P=(2,4)

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Rotate the displayed point by 180° about the origin.

Hint 2

Test the image in the equation.

Worked solution

The image is (−2,−4).
$(-4)^2=16\ne8(-2)$
A centre would preserve every point under that rotation.

The origin is a vertex, not a symmetry centre.

Checks and common pitfalls: Reflection across the x-axis is a different symmetry from a half-turn.

Reasoning checklist · self / teacher assessment

The origin is a vertex, not a symmetry centre.
Reflection across the x-axis is a different symmetry from a half-turn.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

A point on y²=8x has x=6. Find its distance to the focus.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the focus-directrix definition.

Hint 2

The directrix is x=−2.

Worked solution

Compute the horizontal distance to the directrix.
$PF=|6-(-2)|$
The two choices of y share this distance.
$PF=8$

Distance 8.

Checks and common pitfalls: There is no need to select a sign of y.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Classify the equation y²=4ax when a=0 and explain the failed focal interpretation.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Set the parameter to zero explicitly.

Hint 2

The focus would lie on the directrix.

Worked solution

The equation becomes a repeated line.
$y^2=0\Rightarrow y=0$
A nondegenerate parabola requires its focus to lie off the directrix; at a=0 the focus lies on it.

The x-axis as a degenerate conic; not a nondegenerate parabola.

Checks and common pitfalls: A zero coefficient must be treated before dividing by a.

Reasoning checklist · self / teacher assessment

The x-axis as a degenerate conic; not a nondegenerate parabola.
A zero coefficient must be treated before dividing by a.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

A parabola has focus (1,3) and directrix y=−1. Derive its equation and vertex.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The axis is perpendicular to the directrix through the focus.

Hint 2

The vertex bisects the perpendicular focus-directrix gap.

Worked solution

The vertex is (1,1), with a=2 upward.
$(x-1)^2=8(y-1)$
Verify directly by squared distances.
$(x-1)^2+(y-3)^2=(y+1)^2$

Vertex (1,1); (x−1)²=8(y−1).

Checks and common pitfalls: The focus-to-directrix gap is 2a, not a.

Reasoning checklist · self / teacher assessment

Vertex (1,1); (x−1)²=8(y−1).
The focus-to-directrix gap is 2a, not a.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

P lies on y²=8x. Minimize PF+PQ, where F=(2,0) and Q=(8,0). State all equality conditions.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Replace PF by x+2.

Hint 2

Use PQ≥|x−8|, but check equality on the curve.

Worked solution

For x≥0, x+2+|x−8|≥10.
$PF+PQ\ge x+2+|x-8|\ge10$
Equality in the distance bound requires y=0; the parabola then forces x=0.
$P=(0,0),\ PF+PQ=2+8=10$

Minimum 10, attained only at the vertex.

Checks and common pitfalls: The scalar bound is flat for 0≤x≤8, but not every such x is an equality point on the curve.

Reasoning checklist · self / teacher assessment

Minimum 10, attained only at the vertex.
The scalar bound is flat for 0≤x≤8, but not every such x is an equality point on the curve.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Prediction and diagnosis	Diagnose the factor-of-four focus error.
8	Concept derivation	Derive the focus/directrix equation including absolute distance.
10	Guided application	Read rotated and translated parabolas and their focal chords.
12	Transfer and explanation	Solve the distance minimization and analyze its equality set.
5	Exit evidence	Identify the a=0 boundary locus.

Question sequence

Why is distance to x=−a written with an absolute value?
What changes if the focus is left of the vertex?

Board development

Label vertex, focus, directrix and focal chord on the same diagram.
Keep a separate line for the equality case in the distance optimization.

Expected correct and incorrect responses

Students halve 8 instead of dividing it by 4 to locate the focus. Substitute the proposed focus into the defining geometry.
Correct: “4a=8, so the focus is (2,0) and directrix x=−2.”
Incorrect: “The focus is (8,0) because the equation contains 8x.”

Follow-up questions

What set remains when a=0, and why is it excluded from the nondegenerate definition?

Remediation

Compare the vertex-to-focus distance and the vertex-to-directrix distance before expanding squares.

Exit check

For y²=−8x, state both focus and directrix with their signs.

Assessment checklist and sample responses

Credit parameter interpretation, valid squaring and attained equality. Sample: on y²=8x, PF=x+2; the proposed bound is reached at the vertex and nowhere else.
Correct: “4a=8, so the focus is (2,0) and directrix x=−2.”