Focus of a Parabola

Section 2.3 beginning on page 68

The Big IdeasIn this section we will learn about…

• The focus and the directrix of a parabola

• Writing equations for parabolas using the focus, the directrix and the distance formula.

• Recognizing equations of parabolas with a vertex at the origin, identifying the focus and directrix and axis of symmetry using the general formulas, and then graphing them.

• Writing equations for parabolas with a vertex at the origin using the general formulas.

• Standard Equations of parabolas with a vertex at (h,k).• Writing Equations of parabolas with a vertex not on the origin.

Core Vocabulary

Previous

• Perpendicular• Distance Formula• Congruent

New

• Focus• Directrix

The Focus and DirectixA parabola can be defined as the set of all points in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix.

Using the Distance Formula to Write an Equation

Example 1: Use the Distance Formula to write an equation of the parabola with focus and directix .

Based on the definition of a parabola, the line segments drawn from F to P and from P to D are congruent.

𝑃𝐷=𝑃𝐹

𝑫=√(𝒙𝟐−𝒙𝟏)𝟐+(𝒚𝟐− 𝒚𝟏)

𝟐

√(𝑥−𝑥1)2+(𝑦−𝑦 1)

2=√(𝑥−𝑥2)2+(𝑦− 𝑦2)

2

√(𝑥−𝑥)2+(𝑦−(−2))2=√(𝑥−0)2+(𝑦−2)2

√(𝑦+2)2=√𝑥2+(𝑦−2)2

(𝑦+2)2=𝑥2+(𝑦−2)2

𝑦 2+4 𝑦+4=¿𝑥2+ 𝑦2−4 𝑦+48 𝑦=𝑥2

𝑦=18𝑥2

Using the Distance Formula to Write an Equation

𝑃𝐷=𝑃𝐹

√(𝑥−𝑥1)2+(𝑦−𝑦 1)

2=√(𝑥−𝑥2)2+(𝑦− 𝑦2)

2

√(𝑥−𝑥)2+(𝑦−(3))2=√(𝑥−0)2+(𝑦−(−3))2

√(𝑦−3)2=√𝑥2+(𝑦+3)2

(𝑦−3)2=𝑥2+(𝑦+3)2

𝑦 2−6 𝑦+9=¿𝑥2+ 𝑦2+6 𝑦+9−12 𝑦=𝑥2

𝑦=−112

𝑥2

General Equation of a Parabola With Vertex (0,0)

These parabolas have a focus of ) and a directrix of

Graphing an Equation of a ParabolaExample 2: Identify the focus, directrix, and axis of symmetry of Graph the equation.

Step 1: Write in standard form

Step 2: Find focus, directrix, axis of sym

Step 3: Find two points one side of the axis of sym, then reflect them to the other side of the axis of sym.

−4 𝑥=𝑦2

𝑥=−14𝑦2

𝑥=14𝑝

𝑦2

𝑝=−1

Focus at

Directrix

𝑦=1𝑦=2

𝑥=−14𝑦2 𝑥=−

14(1)2 𝑥=−0.25 (−0.25,1)

𝑥=−14𝑦2 𝑥=−

14(2)2 𝑥=−1 (−1,2)

Axis of sym is the x-axis

Writing Equations of Parabolas

Because the vertex is at the origin and the parabola opens down, the general equation is…

The directrix is for this type of parabola is , so …

Substitute that value of p into the general equation..

𝑦=14𝑝

𝑥2

𝑝=−3

𝑦=1

4 (−3)𝑥2 𝑦=

1−12

𝑥2 𝑦=−112

𝑥2

Quick Practice

𝑥=112

𝑦 2 𝑥=−18𝑦2 𝑦=

16𝑥2

Standard Equation of a Parabola With Vertex at

Writing an Equation of a Translated Parabola

Example 4: Write an equation of the parabola shown.

This is a side facing parabola so the standard equation is…

The Vertex is

The Focus is ..

𝑥=14𝑝

(𝑦−𝑘)2+h

(h+𝑝 ,𝑘) (10,2)

h=6 𝑘=2 𝑝=4

From vertex

h+𝑝=10

(6,2)

𝑥=14𝑝

(𝑦−𝑘)2+h

𝑥=14 (4)

(𝑦−2)2+6

𝑥=116

(𝑦−2)2+6

Quick Practice8) Write an equation of a parabola with vertex (−1, 4) and focus (−1, 2).

If the vertex and the focus have the same x-coordinate, I know this is a parabola that opens up or down.

𝑦=14𝑝

(𝑥−h)2+𝑘(h ,𝑘+𝑝 )

Focus =

h=−1𝑘=4 𝑝=−2𝑘+𝑝=2Vertex =

𝑦=14𝑝

(𝑥−h)2+𝑘

𝑦=1

4 (−2)(𝑦−−1)2+4

𝑦=−18(𝑥+1)2+4

4+𝑝=2

(h ,𝑘)

Solving a Real Life Problem

Quick Practice9) A parabolic microwave antenna is 16 feet in diameter. Wire an equation that represents the cross section of the antenna with its vertex at (0,0) and its focus 10 feet to the right of the vertex. What is the depth of the antenna?

With the vertex at the origin and the focus to the right of the vertex, we have a parabola that opens horizontally.

General form of the equation:

Because the focus is 10ft to the right of the vertex:

Specific Equation:

Since the antenna is 16 feet in diameter, we can find the depth by finding the x value when (the distance from the center of the antenna to the outer edge).

𝑥=14𝑝

𝑦2

𝑝=10𝑥=

140

𝑦2

𝑥=140

𝑦2 𝑥=140

(8)2 𝑥=14064 𝑥=1.6 The antenna is

1.6 feet deep.