2.3 Focus of a Parabola EEssentialssential QQuestionuestion What is the focus of a parabola? COMMON CORE Analyzing Satellite Dishes Learning Standards HSF-IF.B.4 Work with a partner. Vertical rays enter a satellite dish whose cross section is a HSF-IF.C.7c parabola. When the rays hit the parabola, they refl ect at the same angle at which they HSG-GPE.A.2 entered. (See Ray 1 in the fi gure.) a. Draw the refl ected rays so that they intersect the y-axis. b. What do the refl ected rays have in common? c. The optimal location for the receiver of the satellite dish is at a point called the focus of the parabola. Determine the location of the focus. Explain why this makes sense in this situation. y Ray Ray Ray 2 incoming angle outgoing y = 1 x2 angle 4 1 CONSTRUCTING VIABLE ARGUMENTS −2 −1 12x To be profi cient in math, you need to make conjectures and build logical progressions of Analyzing Spotlights statements to explore the Work with a partner. Beams of light are coming from the bulb in a spotlight, located truth of your conjectures. at the focus of the parabola. When the beams hit the parabola, they refl ect at the same angle at which they hit. (See Beam 1 in the fi gure.) Draw the refl ected beams. What do they have in common? Would you consider this to be the optimal result? Explain. outgoing y angle = 1 2 2 y 2 x Beam bulb 1 incoming angle Beam −212−1 Beam x CCommunicateommunicate YourYour AnswerAnswer 3. What is the focus of a parabola? 4. Describe some of the properties of the focus of a parabola. Section 2.3 Focus of a Parabola 67 2.3 Lesson WWhathat YYouou WWillill LLearnearn Explore the focus and the directrix of a parabola. Write equations of parabolas. Core VocabularyVocabulary Solve real-life problems. focus, p. 68 directrix, p. 68 Exploring the Focus and Directrix Previous Previously, you learned that the graph of a quadratic function is a parabola that opens perpendicular up or down. A parabola can also be defi ned as the set of all points (x, y) in a plane that Distance Formula are equidistant from a fi xed point called the focus and a fi xed line called the directrix. congruent axis of The focus is in the interior symmetry of the parabola and lies on the axis of symmetry. The directrix is perpendicular to the The vertex lies halfway axis of symmetry. between the focus and the directrix. Using the Distance Formula to Write an Equation STUDY TIP The distance from a Use the Distance Formula to write an equation of the y point to a line is defi ned parabola with focus F(0, 2) and directrix y = −2. F(0, 2) as the length of the P(x, y) perpendicular segment SOLUTION x from the point to the line. Notice the line segments drawn from point F to point P and from point P to point D. By the defi nition of a D(x, −2) parabola, these line segments must be congruent. y = −2 PD = PF Defi nition of a parabola —— —— − 2 + − 2 = − 2 + − 2 √(x x1) (y y1) √(x x2) (y y2) Distance Formula —— —— − 2 + − − 2 = − 2 + − 2 √(x x) (y ( 2)) √(x 0) (y 2) Substitute for x1, y1, x2, and y2. — —— √(y + 2)2 = √x2 + (y − 2)2 Simplify. (y + 2)2 = x2 + (y − 2)2 Square each side. y2 + 4y + 4 = x2 + y2 − 4y + 4 Expand. 8y = x2 Combine like terms. 1 = — 2 y 8 x Divide each side by 8. MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com 1. Use the Distance Formula to write an equation y of the parabola with focus F(0, −3) and D(x, 3) directrix y = 3. y = 3 x P(x, y) F(0, −3) 68 Chapter 2 Quadratic Functions You can derive the equation of a parabola that opens up or down with vertex (0, 0), focus (0, p), and directrix y = −p using the procedure in Example 1. —— —— 2 2 2 2 y √(x − x) + ( y − (−p)) = √(x − 0) + (y − p) F(0, p) (y + p)2 = x2 + (y − p)2 P(x, y) y2 + 2py + p2 = x2 + y2 − 2py + p2 x 4py = x2 D(x, −p) y = −p 1 y = — x2 4p LOOKING FOR The focus and directrix each lie ∣ p ∣ units from the vertex. Parabolas can also open left 1 STRUCTURE or right, in which case the equation has the form x = — y2 when the vertex is (0, 0). 4p 1 Notice that y = — x2 is 4p of the form y = ax2. So, CCoreore CConceptoncept changing the value of Standard Equations of a Parabola with Vertex at the Origin p vertically stretches or Vertical axis of symmetry (x = 0) shrinks the parabola. 1 Equation: y = — x2 y directrix: y 4p focus: y = −p (0, p) Focus: (0, p) vertex: (0, 0) vertex: (0, 0) x x Directrix: y = −p focus: directrix: (0, p) y = −p p > 0 p < 0 Horizontal axis of symmetry (y = 0) 1 Equation: x = — y2 y y 4p directrix: focus: focus: x = −p Focus: (p, 0) (p, 0) (p, 0) vertex: x vertex: x STUDY TIP Directrix: x = −p directrix: x = −p (0, 0) (0, 0) Notice that parabolas opening left or right do not represent functions. p > 0 p < 0 Graphing an Equation of a Parabola Identify the focus, directrix, and axis of symmetry of −4x = y2. Graph the equation. SOLUTION Step 1 Rewrite the equation in standard form. −4x = y2 Write the original equation. 1 x = − — y2 Divide each side by –4. 4 y Step 2 Identify the focus, directrix, and axis of symmetry. The equation has the form 4 1 = x = — y2, where p = −1. The focus is (p, 0), or (−1, 0). The directrix is x 1 4p x = −p, or x = 1. Because y is squared, the axis of symmetry is the x-axis. (−1, 0) −4 −2 2 x Step 3 Use a table of values to graph the y 0 ±1 ±2 ±3 ±4 equation. Notice that it is easier to substitute y-values and solve for x. x 0 −0.25 −1 −2.25 −4 −4 Opposite y-values result in the same x-value. Section 2.3 Focus of a Parabola 69 Writing Equations of Parabolas Writing an Equation of a Parabola y Write an equation of the parabola shown. 4 directrix SOLUTION vertex Because the vertex is at the origin and the axis of symmetry is vertical, the equation −4 4 x 1 has the form y = — x2. The directrix is y = −p = 3, so p = −3. Substitute −3 for p to −2 4p write an equation of the parabola. 1 1 y = — x2 = − — x2 4(−3) 12 = −—1 2 So, an equation of the parabola is y 12 x . MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation. 2. y = 0.5x2 3. −y = x2 4. y2 = 6x Write an equation of the parabola with vertex at (0, 0) and the given directrix or focus. 3 = − − — 5. directrix: x 3 6. focus: ( 2, 0) 7. focus: ( 0, 2 ) The vertex of a parabola is not always at the origin. As in previous transformations, adding a value to the input or output of a function translates its graph. CCoreore CConceptoncept Standard Equations of a Parabola with Vertex at (h, k) Vertical axis of symmetry (x = h) = x = h 1 2 x h Equation: y = — ( x − h) + k y x 4p y (h, k + p) STUDY TIP Focus: (h, k + p) y = k − p The standard form for a Directrix: y = k − p = − vertical axis of symmetry y k p (h, k) (h, k) looks like vertex form. To x (h, k + p) remember the standard form for a horizontal axis p > 0 p < 0 of symmetry, switch x and Horizontal axis of symmetry (y = k) y, and h and k. 1 x = h − p y Equation: x = — ( y − k)2 + h y 4p (h, k) Focus: (h + p, k) (h + p, k) y = k x = − Directrix: x h p y = k (h + p, k) x (h, k) x = h − p p > 0 p < 0 70 Chapter 2 Quadratic Functions Writing an Equation of a Translated Parabola y Write an equation of the parabola shown. 8 SOLUTION 4 vertex focus Because the vertex is not at the origin and the axis of symmetry is horizontal, the 1 equation has the form x = — ( y − k)2 + h. The vertex (h, k) is (6, 2) and the focus 4 12 16 x 4p (h + p, k) is (10, 2), so h = 6, k = 2, and p = 4. Substitute these values to write an equation of the parabola. 1 1 x = — ( y − 2)2 + 6 = — ( y − 2)2 + 6 4(4) 16 1 = — − 2 + So, an equation of the parabola is x 16 ( y 2) 6. Solving Real-Life Problems Parabolic refl ectors have cross Focus Focus sections that are parabolas. Incoming sound, light, or other energy that arrives at a parabolic refl ector parallel to the axis of Diagram 1 Diagram 2 symmetry is directed to the focus (Diagram 1). Similarly, energy that is emitted from the focus of a parabolic refl ector and then strikes the refl ector is directed parallel to the axis of symmetry (Diagram 2). Solving a Real-Life Problem y An electricity-generating dish uses a parabolic refl ector to concentrate sunlight onto a high-frequency engine located at the focus of the refl ector.