Parabola Module PDF

Parabola  Objectives At the end of the lesson, 1 Define parabola. you are expected to: 2 Determine the standard form of the equation of a parabola.  Parabola A parabola is the set of all po...

Parabola  Objectives At the end of the lesson, 1 Define parabola. you are expected to: 2 Determine the standard form of the equation of a parabola.  Parabola A parabola is the set of all points in a plane that are equidistant from a fixed line called a directrix, and a fixed point called focus.  Forms of Equation Standard Form General Form (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) 𝐴𝑥 2 + 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0 (𝑥 − ℎ)2 = −4𝑝(𝑦 − 𝑘) 2 (𝑦 − 𝑘) = 4𝑝(𝑥 − h) 𝐵𝑦 2 + 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0 (𝑦 − 𝑘)2 = −4𝑝(𝑥 − h)  Tanong Ko, Sagot Mo! 1) 25𝑥 2 + 64𝑦 2 − 350𝑥 + 256𝑦 − 119 = 0 Not a Parabola 2) 𝑦 2 + 16𝑥 + 14𝑦 − 31 = 0 Parabola 3) 𝑥 2 − 2𝑥 + 12𝑦 + 61 = 0 Parabola Parts of a Parabola ▪ Focus- fixed point ▪ Directrix- fixed line ▪ Axis of Symmetry- divides the parabola into two parts ▪ Vertex- a point that lies on the axis of symmetry. Parts of a Parabola ▪ Focal Distance (𝒑)- the distance from the vertex to the focus (or directrix) ▪ Latus rectum (4p)- is a segment containing the focus, with endpoints on the parabola and perpendicular to the axis of symmetry.  Tanong Ko, Sagot Mo! Axis of Directrix Symmetry Vertex Latus Rectum Focus  Graph of a Parabola ▪ If the graph opens upward then ▪ If the graph opens downward then the vertex is in the lowest point. the vertex is in the highest point.  Standard Equation of a Parabola Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) Upward (ℎ, 𝑘) (ℎ, 𝑘 + 𝑝) 𝑦 = 𝑘 − 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 + 𝑝) (𝑥 − ℎ)2 = −4𝑝(𝑦 − 𝑘) Downward (ℎ, 𝑘) (ℎ, 𝑘 − 𝑝) 𝑦 = 𝑘 + 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 − 𝑝) (𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ) To the right (ℎ, 𝑘) (ℎ + 𝑝, 𝑘) 𝑥 = ℎ − 𝑝 𝑦 = 𝑘 (ℎ + 𝑝, 𝑘 ± 2𝑝) (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left (ℎ, 𝑘) (ℎ − 𝑝, 𝑘) 𝑥 = ℎ + 𝑝 𝑦 = 𝑘 (ℎ − 𝑝, 𝑘 ± 2𝑝)  Example 1: 𝑥 𝒙𝟐 − 𝟒𝒚 = 𝟎 2 4𝑝 = 4 − 4𝑦 = 0 𝑥 2 = 4𝑦 𝒑=𝟏 Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) Upward (ℎ, 𝑘) (ℎ, 𝑘 + 𝑝) 𝑦 = 𝑘 − 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 + 𝑝) 𝑥 2 = 4𝑦 Upward (0,0) (0,0 + 1) 𝑦 =0−1 𝑥 = 0 (0 ± 2(1), 0 + 1) 𝒙𝟐 = 𝟒𝒚 Upward (𝟎, 𝟎) (𝟎, 𝟏) 𝒚 = −𝟏 𝒙=𝟎 𝟐, 𝟏 , (−𝟐, 𝟏)  k h Example 2: (𝒚 − 𝟑)𝟐 = −𝟏𝟐(𝒙 + 𝟒) −4𝑝 = −12 −4𝑝 −12 = −4 −4 𝒑=𝟑 Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left (ℎ, 𝑘) (ℎ − 𝑝, 𝑘) 𝑥 = ℎ + 𝑝 𝑦 = 𝑘 (ℎ − 𝑝, 𝑘 ± 2𝑝) (𝑦 − 3)2 = −12(𝑥 + 4) To the left (−4,3) (−4 − 3,3) 𝑥 = −4 + 3 𝑦 = 3 (−4 − 3,3 ± 2(3)) (𝒚 − 𝟑)𝟐 = −𝟏𝟐(𝒙 + 𝟒) To the left (−𝟒, 𝟑) (−𝟕, 𝟑) 𝒙 = −𝟏 𝒚=𝟑 −𝟕, 𝟗 , (−𝟕, −𝟑) Tanong Ko, Sagot Mo! 1 1) What is the vertex of the parabola 𝑥 = 2 − 𝑦? 2 2) What is the vertex of the parabola (𝑦 − 8)2 = 7(𝑥 + 15)?  Tanong Ko, Sagot Mo! 3) What is the equation of the directrix of the parabola (𝑥 − 1)2 = 8(𝑦 − 4)? Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) Upward (ℎ, 𝑘) (ℎ, 𝑘 + 𝑝) 𝑦 = 𝑘 − 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 + 𝑝) Directrix: 𝑦 = 𝑘 − 𝑝 4𝑝 = 8 𝑉(ℎ, 𝑘) 𝑦 =4−2 4𝑝 8 𝑉(1,4) = → 𝒑=𝟐 𝒚=𝟐 4 4 𝒌=𝟒 Tanong Ko, Sagot Mo! 1 1) What is the vertex of the parabola 𝑥 = 2 − 𝑦? 2 𝑽 (𝟎, 𝟎) 2) What is the vertex of the parabola (𝑦 − 8)2 = 7(𝑥 + 15)? 𝑽 (−𝟏𝟓, 𝟖)  Tanong Ko, Sagot Mo! 3) What is the equation of the directrix of the parabola (𝑥 − 1)2 = 8(𝑦 − 4)? Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) Upward (ℎ, 𝑘) (ℎ, 𝑘 + 𝑝) 𝑦 = 𝑘 − 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 + 𝑝) Directrix: 𝑦 = 𝑘 − 𝑝 4𝑝 = 8 𝑉(ℎ, 𝑘) 𝑦 =4−2 4𝑝 8 𝑉(1,4) = → 𝒑=𝟐 𝒚=𝟐 4 4 𝒌=𝟒  Tanong Ko, Sagot Mo! 4) What point is the focus of the parabola 𝑦 2 = −10(𝑥 − 4)? Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left (ℎ, 𝑘) (ℎ − 𝑝, 𝑘) 𝑥 = ℎ + 𝑝 𝑦 = 𝑘 (ℎ − 𝑝, 𝑘 ± 2𝑝) 5 𝑉(ℎ, 𝑘) Focus: (ℎ − 𝑝, 𝑘) → (4 − 2 , 0) −4𝑝 = −10 𝑉(4,0) 𝟑 −4𝑝 −10 𝟓 Focus: ( , 𝟎) = → 𝒑= 𝟐 𝒉=𝟒 𝟐 −4 −4 𝒌=𝟎  Tanong Ko, Sagot Mo! −4𝑝 = −12 → −4𝑝 −12 5. 𝑦 2 + 12𝑥 = 0 = −4 −4 𝒚𝟐 = −𝟏𝟐𝒙 𝒑=𝟑 Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left (ℎ, 𝑘) (ℎ − 𝑝, 𝑘) 𝑥 = ℎ + 𝑝 𝑦 = 𝑘 (ℎ − 𝑝, 𝑘 ± 2𝑝) 𝒚𝟐 = −𝟏𝟐𝒙 To the left (0,0) (0 − 3,0) 𝑥 =0+3 𝑦 = 0 (0 − 3,0 ± 2(3)) 𝒚𝟐 = −𝟏𝟐𝒙 To the left (𝟎, 𝟎) (−𝟑, 𝟎) 𝒙=𝟑 𝒚=𝟎 −𝟑, 𝟔 , (−𝟑, −𝟔) Example 3: Reduce this equation of a parabola 𝑥 2 − 6𝑥 − 𝑦 + 11 = 0 to standard form, then identify the parts being asked. 𝑥 2 − 6𝑥 − 𝑦 + 11 = 0 (𝑥 2 −6𝑥 + ____) = 𝑦 − 11 + ____ (𝑏 ÷ 2)2 → (−6 ÷ 2)2 = (−3)2 = 𝟗 (𝑥 2 −6𝑥 + 𝟗) = 𝑦 − 11 + 𝟗 𝟐 𝒙−𝟑 = (𝒚 − 𝟐)  h k 4𝑝 = 1 → 4𝑝 1 𝟐 = 𝒙−𝟑 = (𝒚 − 𝟐) 4 4 𝟏 𝒑= 𝟒 Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑥 − ℎ)2 = 4𝑝(𝑦 − 𝑘) Upward (ℎ, 𝑘) (ℎ, 𝑘 + 𝑝) 𝑦 = 𝑘 − 𝑝 𝑥 = ℎ (ℎ ± 2𝑝, 𝑘 + 𝑝) 𝟐 1 1 1 1 𝒙 − 𝟑 = (𝒚 − 𝟐) Upward (3,2) (3,2 + ) 𝑦 =2− 𝑥 = 3 (3 ± 2( ), 2 + ) 4 4 4 4 𝟐 𝟗 𝟕 𝟕 𝟗 𝟓 𝟗 𝒙 − 𝟑 = (𝒚 − 𝟐) Upward (𝟑, 𝟐) (𝟑, ) 𝒚= 𝒙 = 𝟑 ( 𝟐 , 𝟒 ) , ( 𝟐 , ) 𝟒 𝟒 𝟒 Example 4: Reduce this equation of a 4𝑦 + 8𝑥 − 12𝑦 + 38 = 0 to 2 parabola standard form, then identify the parts being asked. 4𝑦 2 + 8𝑥 − 12𝑦 + 38 = 0 → 4𝑦 2 8𝑥 12𝑦 38 + − + =0 38 4 4 4 4 𝑦2 + 2𝑥 − 3𝑦 + =0 4 2 38 𝑦 − 3𝑦 + ____ = −2𝑥 − + ____ 4 −3 2 𝟗 (𝑏 ÷ 2)2 → (−3 ÷ 2)2 = ( ) = 2 𝟒 𝟗 38 𝟗 𝑦2 − 3𝑦 + = −2𝑥 − + 𝟒 4 𝟒 2 𝟐 3 29 𝟑 𝟐𝟗 𝑦− = −2𝑥 − → 𝒚− = −𝟐 𝒙 + 2 4 𝟐 𝟖  k h 4𝑝 = 1 → 4𝑝 1 𝟑 𝟐 𝟏𝟒𝟑 = 𝒚− = −𝟐 𝒙 − 4 4 𝟐 𝟖 𝟏 𝒑= 𝟒 Endpoints of Equation Opening Vertex Focus Directrix AoS Latus Rectum (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left (ℎ, 𝑘) (ℎ − 𝑝, 𝑘) 𝑥 = ℎ + 𝑝 𝑦 = 𝑘 (ℎ − 𝑝, 𝑘 ± 2𝑝) 143 3 143 1 3 143 1 3 143 1 3 1 (𝑦 − 𝑘)2 = −4𝑝(𝑥 − ℎ) To the left ( , ) ( − , ) 𝑥= + 𝑦= ( − , ± 2( )) 8 2 8 4 2 8 4 2 8 4 2 4 𝟏𝟒𝟑 𝟑 𝟏𝟒𝟏 𝟑 𝟏𝟒𝟓 𝟑 𝟏𝟒𝟏 𝟏𝟒𝟏 (𝒚 − 𝒌)𝟐 = −𝟒𝒑(𝒙 − 𝒉) To the left ( , ) ( , ) 𝒙= 𝒚 = ( , 𝟐) , ( , 𝟏) 𝟖 𝟐 𝟖 𝟐 𝟖 𝟐 𝟖 𝟖  More Examples 5 The vertex is at (4, 2) and the focus is at (2,2) Vertex: 𝟒, 2 ℎ = 4, 𝑘 = 2 𝑝 = 4−2 Focus: (𝟐, 2) 𝒑=𝟐 Opening: to the left 2 𝑦−𝑘 = −4𝑝 𝑥 − ℎ 2 𝑦−2 = −4 2 𝑥 − 4 𝟐 𝒚−𝟐 = −𝟖 𝒙 − 𝟒  Tanong Ko, Sagot Mo! The vertex is at (2, 1) and the focus is at (2,4) Vertex: 2, 𝟏 ℎ = 2, 𝑘 = 1 𝑝 = 1−4 Focus: (2, 𝟒) 𝒑=𝟑 Opening: Upward 2 𝑥−ℎ = 4𝑝 𝑦 − 𝑘 2 𝑥−2 = 4(3) 𝑦 − 1 2 𝑥−2 = 12 𝑦 − 1  More Examples 6 The vertex is at (−1,4) and the directrix is x = −5. Directrix: x = −5 Vertex: −1,4 → ℎ = −1, 𝑘 = 4 𝑝 = −1 − (−5) Opening: to the right = −1 + 5 2 y−𝑘 = 4𝑝 x − ℎ 𝒑=𝟒 2 y−4 = 4(4) x + 1) 2 y−4 = 16 x + 1  The vertex is at (4, −4) and the directrix is 𝑦 + 5 = 0 Directrix: 𝑦 = −5 Vertex: 4, −4 ℎ = 4, 𝑘 = −4 𝑝 = −4 − (−5) Opening: Upward 𝑝 = −4 + 5 𝒑=𝟏 2 𝑥−ℎ = 4𝑝 𝑦 − 𝑘 2 𝑥−4 = 4 1 (𝑦 + 4) 2 𝑥−4 =4 𝑦+4 Tanong Ko, Sagot Mo!  3 7 The parabola passes through the point (− , −1), −1,5 and 2 its vertex is at (0, −4) with a horizontal axis of symmetry. Vertex: (0, −4) 2 81 3 𝑦 − −4 =4 − 𝑥−0 Points : (− , −1), −1,5 4 2 Using one of the points of the parabola (-1,5) and its 𝑦+4 2 = −81𝑥 vertex (𝟎, −𝟒), solve p using the equation 𝑦−𝑘 2 = 4𝑝 𝑥 − ℎ 𝑦 2 + 8𝑦 + 16 = −81𝑥 2 5 − −4 = 4𝑝 −1 − 0 𝐲 𝟐 + 𝟖𝐲 + 𝟖𝟏𝐱 + 𝟏𝟔 = 𝟎 81 = −4𝑝 𝑝=− 81 4 More Examples  Problem-Solving 1) A satellite dish shaped like a paraboloid, has a diameter of 6 ft and a depth of 2.5 ft. If the receiver is placed at the focus, how far should the receiver be from the vertex?  Problem-Solving 1) A satellite dish shaped like a paraboloid, has a diameter of 6 ft and a depth of 2.5 ft. If the receiver is placed at the focus, how far should the receiver be from the vertex? 𝑥 2 = 4p𝑦 (3)2 = 4p(2.5) Focus: (0,p) 9 = 10p 9 Focus: (0,0.9) p= or 0.9 10 Therefore, the receiver is 0.9 ft away from the vertex. 2) Water flowing from a suspended garden hose to the ground follows a parabolic path, with the hose's opening as the vertex. Suppose this opening is 10 feet high and that the water strikes the ground a horizontal distance of 8 feet from where the opening is located. At what height is the horizontal distance of the water 1 foot from the opening? 2) Water flowing from a suspended garden hose to the ground follows a parabolic path, with the hose's opening as the vertex. Suppose this opening is 10 feet high and that the water strikes the ground a horizontal distance of 8 feet from where the opening is located. At what height is the horizontal distance of the water 1 foot from the opening? 2 2 8 2 −32y 𝑥 = −4𝑎𝑦 (8) = −4𝑎(−10) 𝑥 2 = −4( )𝑦 (1) = 5 5 Focus: (0, −𝑎) 64 = 40𝑎 −32y 5 = −32𝑦 64 8 𝑥2 = 𝑎 = or 5 y= −5 40 5 32 Let x=1 𝐲 = −𝟎. 𝟏𝟔  Assignment Directions: Answer the following questions on your university notebook. Show your complete solutions. 1. (𝑦 − 8)2 = 12(𝑥 − 4) 2. 𝑦 2 = −5 𝑥 − 8 3. (𝑥 − 4)2 = 12(𝑦 + 8)  Assignment Directions: Write the equation of the parabola in standard form satisfying the prescribed conditions. Show your complete solutions. 4. Page 31, Letter A, no. 1 5. Vertex (-1, 3); Directrix: y=1  Assignment Directions: Reduce the following equations to standard form, then identify the opening, vertex, focus, directrix, axis of symmetry, and endpoints of the latus rectum. 6. 𝑦 2 + 3𝑥 = 0 7. Page 31, Letter B, no. 7  Assignment Directions: Answer the question on your university notebook. Show your illustration and complete solutions. 8. A satellite dish is shown. Since radio signals (parallel to the axis) will bounce off the surface of the dish to the focus, the receiver should be placed at the focus. How far should the receiver be from the vertex, if the dish is 22 ft across, and 7 ft deep at the vertex?