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**8.2 Graph and Write Equations of Parabolas**

Where is the focus and directrix compared to the vertex? How do you know what direction a parabola opens? How do you write the equation of a parabola given the focus/directrix? What is the general equation for a parabola?

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Parabolas A parabola is defined in terms of a fixed point, called the focus, focus and a fixed line, called the directrix. A parabola is the set of all points P(x,y) in the plane whose distance to the focus equals its distance to the directrix. directrix axis of symmetry

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Horizontal Directrix Standard Equation of a parabola with its vertex at the origin is x y D(x, –p) P(x, y) F(0, p) y = –p O x2 = 4py p > 0: opens upward p < 0: opens downward focus: (0, p) directrix: y = –p axis of symmetry: y-axis

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Vertical Directrix Standard Equation of a parabola with its vertex at the origin is x y D(x, –p) P(x, y) F(p, 0) x = –p O y2= 4px p > 0: opens right p < 0: opens left focus: (p, 0) directrix: x = –p axis of symmetry: x-axis

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**Example 1 Graph . Label the vertex, focus, and directrix. y2 = 4px**

Identify p. -4 -2 2 4 y2 = 4(1)x So, p = 1 Since p > 0, the parabola opens to the right. Vertex: (0,0) Focus: (1,0) Directrix: x = -1

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**Example 1 Graph . Label the vertex, focus, and directrix. Y2 = 4x y x**

Use a table to sketch a graph -4 -2 2 4 y x 2 4 -2 -4 1 4 1 4

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**Graph x = –⅛y 2. Identify the focus, directrix, and axis**

of symmetry. SOLUTION STEP 1 Rewrite the equation in standard form. 18 x = – Write original equation. – 8x = y 2 Multiply each side by – 8. STEP 2 Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = – 2. The focus is (p, 0), or (– 2, 0). The directrix is x = – p, or x = 2. Because y is squared, the axis of symmetry is the x - axis.

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STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values.

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**Rewrite the equation in standard form. Y 2 = 4 (– )x 3 2 STEP 2**

Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola. 1. Y 2 = –6x SOLUTION STEP 1 Rewrite the equation in standard form. Y 2 = 4 (– )x 3 2 STEP 2 Identify the focus, directrix, and axis of symmetry. The equation has the form y 2 = 4px where p = – The focus is (p, 0), or (– , 0). The directrix is x = – p, or x = . Because y is squared, the axis of symmetry is the x - axis. 3 2

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STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values. 2.45 4.24 4.90 5.48

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**Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.**

14 3. y = – x 2 SOLUTION STEP 1 Rewrite the equation in standard form. y = – x 2 14 Write original equation. Multiply each side by – 4. – 4y = x 2 equation STEP 2 focus directrix axis of symmetry x 2 = – 4 0, –1 y = 1 Vertical x = 0

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STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative y - values. y x 2 4.47 2.83 3.46 4

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**Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.**

4. x = – y 2 13 SOLUTION STEP 1 Rewrite the equation in standard form. 13 x = Y 2 Write original equation. Multiply each side by 3. 3x = y 2 equation STEP 2 focus directrix axis of symmetry Y 2 = x 3 4 0, 3 4 x = – 3 4 Horizontal y = 0

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STEP 3 Draw the parabola by making a table of values and plotting points. Because p < 0, the parabola opens to the left. So, use only negative x - values. y x 1.73 3.87 2.45 3 3.46

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Example 2 Write the standard equation of the parabola with its vertex at the origin and the directrix y = -6. Since the directrix is below the vertex, the parabola opens up Since y = -p and y = -6, p = 6 x2=4(6)y x2 = 24y

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**Write an equation of the parabola shown.**

SOLUTION The graph shows that the vertex is (0, 0) and the directrix is y = – p = for p in the standard form of the equation of a parabola. 3 2 – x2 = 4py Standard form, vertical axis of symmetry x2 = 4• ( )y 3 2 Substitute for p 3 2 x 2 = 6y Simplify.

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Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus. 5. Directrix: y = 2 SOLUTION x 2 = 4py Standard form, vertical axis of symmetry x 2 = 4 (–2)y Substitute –2 for p x 2 = – 8y Simplify.

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Write the standard form of the equation of the parabola with vertex at (0, 0) and the given directrix or focus. 8. Focus: (0, 3) SOLUTION x 2 = 4py Standard form, vertical axis of symmetry x 2 = 4 (3)y Substitute 3 for p x 2 = 12y Simplify.

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Solar Energy The EuroDish, developed to provide electricity in remote areas, uses a parabolic reflector to concentrate sunlight onto a high-efficiency engine located at the reflector’s focus. The sunlight heats helium to 650°C to power the engine. • Write an equation for the EuroDish’s cross section with its vertex at (0, 0). • How deep is the dish? Read more on page 498

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**Where is the focus and directrix compared to vertex?**

The focus is a point on the line of symmetry and the directrix is a line below the vertex. The focus and directrix are equidistance from the vertex. How do you know what direction a parabola opens? x2, graph opens up or down, y2, graph opens right or left How do you write the equation of a parabola given the focus/directrix? Find the distance from the focus/directrix to the vertex (p value) and substitute into the equation. What is the general equation for a parabola? x2= 4py (opens up [p>0] or down [p<0]), y2 = 4px (opens right [p>0] or left [p<0])

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8.2 Assignment p. 499, odd, 27-33 odd, 39-45 odd

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**8.2 Graph and Write Equations of Parabolas day 2**

What does it mean if a parabola has a translated vertex? What general equations can you use for a parabola when the vertex has been translated?

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**Standard Equation of a Translated Parabola**

Vertical axis: (x − h)2 = 4p(y − k) vertex: (h, k) focus: (h, k + p) directrix: y = k – p axis of symmetry: x = h

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**Standard Equation of a Translated Parabola**

Horizontal axis: (y − k)2 = 4p(x − h) vertex: (h, k) focus: (h + p, k) directrix: x = h - p axis of symmetry: y = k

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Example 3 Write the standard equation of the parabola with a focus at F(-3,2) and directrix y = 4. Sketch the info. The parabola opens downward, so the equation is of the form (x − h)2 = 4p(y − k) vertex: (-3,3) h = -3, k = 3 p = -1 (x + 3)2 = 4(−1)(y − 3)

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**Draw a curve through the points.**

Graph (x – 2)2 = 8 (y + 3). SOLUTION STEP 1 Compare the given equation to the standard form of an equation of a parabola . You can see that the graph is a parabola with vertex at (2, – 3) ,focus (2, – 1) and directrix y = – 5 Draw the parabola by making a table of value and plot y point. Because p > 0, he parabola open to the right. So use only points x- value STEP 2 x 1 2 3 4 5 y –2.875 –3 – 2.875 –2.5 – 1.875 STEP 3 Draw a curve through the points.

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Example 4 Write an equation of a parabola whose vertex is at (−2,1) and whose focus is at (−3, 1). Begin by sketching the parabola. Because the parabola opens to the left, it has the form (y −k)2 = 4p(x − h) Find h and k: The vertex is at (−2,1) so h = −2 and k = 1 Find p: The distance between the vertex (−2,1) and the focus (−3,1) by using the distance formula. p = −1 (y − 1)2 = −4(x + 2)

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**Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3.**

Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3). SOLUTION STEP 1 Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (y – k)2 = 4p(x – h) where p < 0. STEP 2 Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3. STEP 3 Find p. The vertex (– 2, 3) and focus (4, 3) both lie on the line y = 3, so the distance between them is p | = | – 4 – (– 2) | = 2, and thus p = +2. Because p < 0, it follows that p = – 2, so 4p = – 8. The standard form of the equation is (y – 3)2 = – 8(x + 2).

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**Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1.**

Write the standard form of a parabola with vertex at (3, – 1) and focus at (3, 2). SOLUTION STEP 1 Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (x – h)2 = 4p(y – k) where p > 0. STEP 2 Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1. STEP 3 Find p. The vertex (3, – 1) and focus (3, 2) both lie on the line x = 3, so the distance between them is p | = | – 2 – (– 1) | = 3, and thus p = + 3. Because p > 0, it follows that p = 3, so 4p = 12. The standard form of the equation is (x – 3)2 = 12(y + 1)

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**What does it mean if a parabola has a translated vertex?**

It means that the vertex of the parabola has been moved from (0,0) to (h,k). What general equations can you use for a parabola when the vertex has been translated? (y-k)2 =4p(x-h) (x-h)2 =4p(y-k)

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**8.2 Assignment day 2 p. 499, 26-32 even, 40-46 even**

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