Conics D Wetzel 2009 Parabolas Parabola the set

Conics D. Wetzel 2009

Parabolas Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix. The cross section of a headlight is an example of a parabola. . . Directrix The light source is the Focus

Here are some other examples of the parabola. . .

d 2 d 1 Focus d 1 Vertex d 2 d 3 Directrix Notice that the vertex is located at the midpoint between the focus and the directrix. . . Also, notice that the distance from the focus to any point on the parabola is equal to the distance from that point to the directrix. . . We can determine the coordinates of the focus, and the equation of the directrix, given the equation of the parabola. .

Standard Equation of a Parabola: (Vertex at the origin) Equation 2 ax = y Focus Directrix (0, 1/4 a) y = -1/4 a (If the x term is squared, the parabola goes up or down) Equation 2 ay = x Focus Directrix (1/4 a, 0) x = – 1/4 a (If the y term is squared, the parabola goes left or right)

Example: Determine the focus and directrix of the 2 parabola y = 4 x : Since x is squared, the parabola goes up or down… 2 and the equation is: ax = y To find the focus, Let P represent the distance from the vertex to the focus p = 1/4 a p = 1/(4)(4); p = 1/16 Focus: (0, p) Directrix: y = –p Focus: (0, 1/16) Directrix: y = – 1/16 See what this parabola looks like. . .

Example: Determine the focus and directrix of the 2 parabola – 3 y – 12 x = 0 : Since y is squared, the parabola goes left or right… 2 The equation is: ay = x Solve for x: -3 y 2 = 12 x -3/12 y 2 = 12/12 x -1/4 y 2 = x p = 1/4 a so, p = 1/4(-1/4) = -1 Focus: (p, 0) Directrix: x = –p Focus: (– 1, 0) Directrix: x = 1 See what this parabola looks like. . .

Examples: Write the standard form of the equation of the parabola with focus at (0, 3) and vertex at the origin. Since the focus is on the y axis, (and vertex at the origin) the parabola goes up or down… The equation is: ax 2 = y Since p = 3, 1/4 a = 3 and a = 1/12 The standard form of the equation is: 1/12 x 2 = y

Example: Write the standard form of the equation of the parabola with directrix x = – 1 and vertex at the origin. Since the directrix is parallel to the y axis, (and vertex at the origin) the parabola goes left or right… The standard equation is: ay 2 = x Since p = 1, 1/4 a = 1 and a = ¼ The standard form of the equation is: 1/4 y 2 = x

Circles D. Wetzel 2009

circle: in a plane, the set of points equidistant from a given point, called the center. radius: any segment whose endpoints are the center and a point on the circle. If the circle is centered at (0, 0), and the radius is r, then the distance to any point, (x, y) on the circle (using the distance formula) is square both sides. . . now simplify. . . This is the standard form of a circle with center (0, 0) and radius r. (x, y)

Example: Write an equation of the circle with its center at the origin with the point (– 3, 4) on the circle. Use the Standard Equation of the Circle to find the radius of the circle. . .

Ellipses D. Wetzel 2009

Ellipses Ellipse: set of all points in a plane such that the sum of the distances from two given points in a plane, called the foci, to any point on the curve is the same. co-vertex Sum of the distances: foci 12 units vertex co-vertex The major axis is the line segment joining the vertices (through the foci) The minor axis is the line segment joining the co-vertices (perpendicular to the major axis)

Standard Equation of an Ellipse (Center at Origin) This is the equation if the major axis is horizontal. (0, b) (–a, 0) (–c, 0) (a, 0) (0, –b) The foci of the ellipse lie on the major axis, c units from the center, where c 2 = a 2 – b 2

Standard Equation of an Ellipse (Center at Origin) (0, b) This is the equation if the major axis is vertical. (notice it looks the same) (0, c) (–a, 0) (0, –c) (0, –b) The foci of the ellipse lie on the major axis, c units from the center, where c 2 = a 2 – b 2

Example: Write an equation of the ellipse whose vertices are (0, – 3) and (0, 3) and whose co-vertices are (– 2, 0) and (2, 0). Find the foci of the ellipse. Since the major axis is (0, 3) (0, c) (– 2, 0) (a, 0) (0, –c) vertical, the equation is the following: Since b = 3 a = 2 The equation is (0, – 3) Use c 2 = b 2 – a 2 to find c. c 2 = 32 – 22 c 2 = 9 – 4 = 5 c= The foci are

Example: Write the equation in standard form of 9 x 2 + 16 y 2 = 144. Find the foci and vertices of the ellipse. Get the equation in standard form by dividing by 144: Simplify. . . That means a = 4 b=3 (0, 3) (– 4, 0) Use c 2 = a 2 – b 2 to find c. c 2 = 42 – 32 c 2 = 16 – 9 = 7 c= (4, 0) (–c, 0) (0, -3) Vertices: Foci: Center: (0, 0)

Hyperbolas D. Wetzel 2009

Hyperbolas Hyperbola: set of all points such that the difference of the distances from any point to the foci is constant. foci vertices asymptotes

Standard Equation of a Hyperbola (Center at Origin) (0, b) (–c, 0) (–a, 0) This is the equation if the transverse axis is horizontal. (notice it looks a lot like the equation of an ellipse. ) (c, 0) (a, 0) (0, –b) The foci of the hyperbola lie on the major axis, c units from the center, where c 2 = a 2+ b 2

Standard Equation of a Hyperbola (Center at Origin) (0, c) (–a, 0) (0, b) This is the equation if the transverse axis is vertical. (Notice that y is first for vertical and x was first for horizontal. ) (a, 0) (0, –b) (0, –c) The foci of the hyperbola lie on the major axis, c units from the center, where c 2 = a 2+ b 2

2 2 Example: Write the equation in standard form of 4 x – 16 y = 64. Find the foci and vertices of the hyperbola. Divide both side of the equation by 64 to get the standard form: Simplify. . . That means a = 4 b=2 (0, 2) (– 4, 0) Use c 2 = a 2 + b 2 to find c. c 2 = 42 + 22 c 2 = 16 + 4 = 20 c= (4, 0) (c, 0) (–c, 0) (0, -2) Vertices: Foci:

Example: Write an equation of the hyperbola whose foci are (0, – 6) and (0, 6) and whose vertices are (0, – 4) and (0, 4). Its center is (0, 0). Since the major axis is vertical, the equation is the following: (0, 6) (–a, 0) (0, 4) (0, – 6) (a, 0) Since b = 4 and c = 6 , find a. . . 2 2 2 c =b +a 62 = 42 + a 2 36 = 16 + a 2 20 = a 2 The equation of the hyperbola:

How do you graph a hyperbola? To graph a hyperbola, you need to know the center, the vertices, the co-vertices, and the asymptotes. . . The asymptotes intersect at the center of the hyperbola and pass through the corners of a rectangle with corners (± a, ±b) Example: Graph the hyperbola a=4 (– 5, 0) (0, 3) (– 4, 0) (0, -3) b=3 c=5 Draw a rectangle using +a and +b as the sides. . . Draw the asymptotes (The asymptotes are just the diagonals through the box generated by a (5, 0) and b. ). . . Draw the hyperbola. . .

WRITING AND GRAPHING EQUATIONS OF CONICS OFEQUATIONS RATIONAL FUNCTIONS STANDARDGRAPHS FORM OF OF TRANSLATED CONICS In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics. (x – h) 2 + (y – k) 2 = r 2 CIRCLE Horizontal PARABOLA ELLIPSE 2 2 a(y – k) + h = x (x – h) a 2 + 2 2 HYPERBOLA Vertical (y – k) b a(x – h) + k = y 2 2 (x – h) (y – k) – 2 2 a b =1 a>b 2 =1 (x – h) a 2 (y – k) b 2 2 + (y – k) b 2 – 2 (x – h) a 2 2 =1 b>a 2 =1

Writing an Equation of a Translated Parabola Write an equation of the parabola whose vertex is at (– 2, 1) and whose focus is at (– 3, 1). SOLUTION Plot what you have so far: Because the parabola opens to the left, it has the form: x = a(y – k)2 + h Find h and k: The vertex is at (– 2, 1), so h = – 2 and k = 1. Giving you: x = a(y – 1)2 + -2 (– 2, 1)

Writing an Equation of a Translated Parabola Write an equation of the parabola whose vertex is at (– 2, 1) and whose focus is at (– 3, 1). SOLUTION Find p: The distance between the vertex (– 2, 1), and the focus (– 3, 1) is p = (– 2, 1) (– 3 – (– 2)) 2 + (1 – 1) 2 = 1 so p = 1 or p = – 1. Since p < 0, p = – 1. p = 1/4 a so a = -4 The standard form of the equation is x = -4(y – 1) 2 – 2.

Graphing the Equation of a Translated Circle Graph (x – 3) 2 + (y + 2) 2 = 16. SOLUTION Compare the given equation to the standard form of the equation of a circle: (x – h) 2 + (y – k) 2 = r 2 You can see that the graph will be a circle with center at (h, k) = (3, – 2)

Graphing the Equation of a Translated Circle Graph (x – 3) 2 + (y + 2) 2 (3, 2) = 16. SOLUTION r The radius is r = 4 Plot several points that are each 4 units from the center: (3 + 4, – 2 + 0) (3 – 4, – 2 + 0) (3 + 0, – 2 + 4) (3 + 0, – 2 – 4) = = (7, – 2) (– 1, – 2) (3, – 6) Draw a circle through the points. (– 1, – 2) (3, – 6) (7, – 2)

Writing an Equation of a Translated Ellipse Write an equation of the ellipse with foci at (3, 5) and (3, – 1) and vertices at (3, 6) and (3, – 2). SOLUTION Plot the given points and make a rough sketch. The ellipse has a vertical major axis, so its equation is of the form: (x – h) 2 + (y – k) 2 = 1 a 2 b 2 Find the center: The center is halfway between the vertices. (h, k) = (3 + 3 , 6 + ( – 2) = (3, 2) 2 2 (3, 6) (3, 5) (3, – 2) (3, – 1)

Writing an Equation of a Translated Ellipse Write an equation of the ellipse with foci at (3, 5) and (3, – 1) and vertices at (3, 6) and (3, – 2). SOLUTION Find b: The value of b is the distance between the vertex and the center. b = (3 – 3) 2 + (6 – 2) 2 = 0 + 4 2 = 4 Find c: The value of c is the distance between the focus and the center. c = (3 – 3) 2 + (5 – 2) 2 = 0 + 3 2 = 3 (3, 6) (3, 5) (3, – 2) (3, – 1)

Writing an Equation of a Translated Ellipse Write an equation of the ellipse with foci at (3, 5) and (3, – 1) and vertices at (3, 6) and (3, – 2). SOLUTION (3, 6) Find a: Substitute the values of a and c into the equation a 2 = b 2 – c 2. a 2 = 42 – 32 a 2 = 7 a= 7 (3, 5) (3, – 2) 2 (x – 3 ) The standard form is + 7 (y – 2) 2 16 =1 (3, – 1)

Graphing the Equation of a Translated Hyperbola Graph (y + 1) 2 (x + 1) 2 – = 1. 4 (– 1, 0) SOLUTION The y 2 -term is positive, so the transverse axis is vertical. Since a 2 = 4 and b 2 = 1, you know that a = 2 and b = 1. (– 1, – 1) (– 1, – 2) Plot the center at (h, k) = (– 1, – 1). Plot the vertices 1 unit above and below the center at (– 1, 0) and (– 1, – 2). Draw a rectangle that is centered at (– 1, – 1) and is 2 a = 2 units high and 2 b = 4 units wide.

Graphing the Equation of a Translated Hyperbola Graph (y + 1) 2 (x + 1) 2 – = 1. 4 (– 1, 0) SOLUTION The y 2 -term is positive, so the transverse axis is vertical. Since a 2 = 4 and b 2 = 1, you know that a = 2 and b = 1. (– 1, – 1) (– 1, – 2) Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

Here all of the conic section equations in standard form: parabola: 2 y = a(x – h) + k x = a(y – k)2 + h 2 2 ellipse: (x – h) + (y – k) = 1 a 2 b 2 2 2 circle: (x – h) + (y – k) = r hyperbola: 2 (x – h)2 – (y – k)2 = 1 a 2 b 2 (y – k)2 – (x – h)2 = 1 b 2 a 2

How do you write an equation in standard form? Example: Write the equation of the ellipse in standard form 2 x 2 + 3 y 2 + 4 x + 12 y – 10 = 0 Group the x’s and y’s together. . . 2 x 2 + 4 x +3 y 2 + 12 y = 10 Factor out the GCF’s. . . 2(x 2 + 2 x +1 ) + 3(y 2 + 4 y +4) = 10 Complete the square for each +2 +12 variable. What will make each a perfect square trinomial? Add the “real” amount to the other side (remember that they are being distributed) Rewrite as the squares of binomials. . . 2(x + 1)2 + 3(y + 2)2 = 24 Divide to set the right side equal to 1. . . 2(x + 1)2 + 3(y + 2)2 = 24 24 (x +1)2 + (y +2)2 = 1 12 8

Let’s try another…with graphing. Example: Graph the hyperbola 9 x 2 – 4 y 2 + 18 x + 16 y – 43 = 0 Group the x’s and y’s together. . . 2 2 9 x + 18 x – 4 y + 16 y = 43 Factor out the 9(x 2 + 2 x +1 ) – 4(y 2 – 4 y +4) = 43 GCF’s. . . + 9 – 16 Complete the squares. . . 2 2 9(x + 1) – 4(y – 2) =36 9(x + 1)2 – 4(y – 2)2 = 36 36 2 2 (x +1) – (y – 2) = 1 4 9 Center (– 1, 2) a=2 b=3