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Conics

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. The light source is the Focus. Directrix.

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Conics

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  1. Conics D.Wetzel 2009

  2. 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... The light source is the Focus Directrix

  3. Here are some other examples of the parabola...

  4. d2 d1 Focus d2 d3 d1 d3 Vertex 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....

  5. (0, 1/4a) y = -1/4a (If the x term is squared, the parabola goes up or down) Standard Equation of a Parabola:(Vertex at the origin) Equation Focus Directrix ay2 = x (1/4a, 0) x = –1/4a Equation Focus Directrix ax2 = y (If the y term is squared, the parabola goes left or right)

  6. Example: Determine the focus and directrix of the parabola y = 4x2 : Since x is squared, the parabola goes up or down… and the equation is:ax2 = y To find the focus, Let P represent the distance from the vertex to the focus p = 1/4a 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...

  7. Example: Determine the focus and directrix of the parabola –3y2 – 12x = 0 : Since y is squared, the parabola goes left or right… The equation is: ay2 = x Solve for x: -3y2 = 12x -3/12y2 = 12/12x -1/4y2 = x p = 1/4a 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...

  8. 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: ax2 = y Since p = 3, 1/4a = 3 and a = 1/12 The standard form of the equation is:1/12x2 = y

  9. 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: ay2 = x Since p = 1, 1/4a = 1 and a = ¼ The standard form of the equation is:1/4y2 = x

  10. Circles D.Wetzel 2009

  11. 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 isr, then the distance to any point, (x, y) on the circle (using the distance formula) is (x,y) square both sides... now simplify... This is the standard form of a circle with center (0,0) and radius r.

  12. 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...

  13. Ellipses D.Wetzel 2009

  14. Ellipses Sum of the distances: 12 units co-vertex foci vertex vertex 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 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)

  15. This is the equation if the major axis is horizontal. (0, b) Standard Equation of an Ellipse (Center at Origin) (–c, 0) (c, 0) (–a, 0) (a, 0) (0, –b) The foci of the ellipse lie on the major axis, c units from the center, where c2 = a2 – b2

  16. This is the equation if the major axis is vertical. (notice it looks the same) (0, b) Standard Equation of an Ellipse (Center at Origin) (0, c) (–a, 0) (a, 0) (0, –c) (0, –b) The foci of the ellipse lie on the major axis, c units from the center, where c2 = a2 – b2

  17. Since the major axis is vertical, the equation is the following: (0, 3) (0, c) 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. (–2, 0) (a, 0) Since b = 3a = 2 The equation is (0, –c) (0, –3) Use c2 = b2 – a2 to find c. c2 = 32 – 22 c2 = 9 – 4 = 5 c = The foci are

  18. Get the equation in standard form by dividing by 144: Simplify... Example: Write the equation in standard form of 9x2 + 16y2 = 144. Find the foci and vertices of the ellipse. That means a = 4b = 3 Use c2 = a2 – b2 to find c. c2 = 42 – 32 c2 = 16 – 9 = 7 c = (0, 3) (–4,0) (4, 0) Vertices: Foci: Center: (0,0) (–c,0) (c, 0) (0,-3)

  19. Hyperbolas D.Wetzel 2009

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

  21. This is the equation if the transverse axis is horizontal. (notice it looks a lot like the equation of an ellipse.) Standard Equation of a Hyperbola (Center at Origin) (0, b) (–c, 0) (c, 0) (–a, 0) (a, 0) (0, –b) The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2

  22. This is the equation if the transverse axis is vertical. (Notice that y is first for vertical and x was first for horizontal.) (0, c) Standard Equation of a Hyperbola (Center at Origin) (0, b) (–a, 0) (a, 0) (0, –b) (0, –c) The foci of the hyperbola lie on the major axis, c units from the center, where c2 = a2+ b2

  23. Simplify... Example: Write the equation in standard form of 4x2 – 16y2 = 64. Find the foci and vertices of the hyperbola. Divide both side of the equation by 64 to get the standard form: That means a = 4b = 2 Use c2 = a2 + b2 to find c. c2 =42 +22 c2 = 16 + 4 = 20 c = (0, 2) (–4,0) (4, 0) Vertices: Foci: (c, 0) (–c,0) (0,-2)

  24. Since the major axis is vertical, the equation is the following: (0, 6) 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). (0, 4) (–a, 0) (a, 0) Since b = 4andc = 6 , find a... c2=b2+a2 62 =42 +a2 36 = 16 + a2 20 = a2 The equation of the hyperbola: (0, –4) (0, –6)

  25. 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) How do you graph a hyperbola? Example: Graph the hyperbola a = 4b = 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 and b.)... (0, 3) (–5,0) (5, 0) (–4,0) (4, 0) (0,-3) Draw the hyperbola...

  26. GRAPHS OF RATIONAL FUNCTIONS STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS Horizontal Vertical a(y – k)2+ h =x a(x – h)2+ k =y PARABOLA (x – h)2(y – k)2 (x – h)2(y – k)2 + = 1 + = 1 ELLIPSE a2 b2 a > b a2 b2 b > a (x – h)2(y – k)2 (y – k)2(x – h)2 – = 1 – = 1 HYPERBOLA a2 b2 b2 a2 WRITING AND GRAPHING EQUATIONS OF CONICS In the following equations the point (h, k) is the vertexof the parabola and the center of the other conics. (x – h)2 + (y – k)2 = r2 CIRCLE

  27. Writing an Equation of a Translated Parabola (–2, 1) 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

  28. Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is (–3, 1) p=(–3 – (–2))2+ (1 – 1)2= 1 sop = 1 or p= – 1. Since p < 0, p =– 1. p = 1/4a so a = -4 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 (–2, 1) The standard form of the equation is x = -4(y – 1)2 – 2 .

  29. 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: (3, – 2) (x – h)2+ (y – k)2= r2 You can see that the graph will be a circle with center at (h, k) = (3, – 2).

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

  31. (3, 6) (3, 5) The ellipse has a vertical major axis, so its equation is of the form: (x – h)2(y – k)2 + = 1 a2 b2 Find the center: The center is halfwaybetween the vertices. (3, –2) (3, –1) (3 + 3 6 + ( –2) (h, k) = , = (3, 2) 2 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.

  32. Writing an Equation of a Translated Ellipse (3, 6) (3, 5) b = (3 – 3)2 + (6 – 2)2= 0 + 42= 4 (3, –2) (3, –1) c= (3 – 3)2 + (5 – 2)2= 0 + 32= 3 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 distancebetween the vertex and the center. Find c: The value of c is the distancebetween the focus and the center.

  33. Writing an Equation of a Translated Ellipse (3, 6) (3, 5) a2 = 42–32 a2 = 7 a = 7 (3, –2) (3, –1) (x – 3)2(y – 2)2 The standard form is + = 1 7 16 Write an equation of the ellipse with foci at (3,5) and (3, –1) and vertices at (3,6) and (3, –2). SOLUTION Find a: Substitute the values of aandc into the equation a2 = b2–c2.

  34. Graphing the Equation of a Translated Hyperbola (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 4 and b2 = 1, you know thata = 2 and b = 1. 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 2a = 2 units high and 2b = 4 units wide.

  35. Graphing the Equation of a Translated Hyperbola (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 4 and b2 = 1, you know thata = 2 and b = 1. Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

  36. circle: (x – h)2 + (y – k)2 = r2 Here are all of the conic section equations in standard form: hyperbola:(x – h)2 – (y – k)2 = 1 a2 b2 (y – k)2 – (x – h)2 = 1 b2 a2 ellipse:(x – h)2 + (y – k)2 = 1 a2 b2 parabola: y = a(x – h)2 + k x = a(y – k)2 + h

  37. Example: Write the equation of the ellipse in standard form2x2 + 3y2 + 4x + 12y – 10 = 0 Group the x’s and y’s together... 2x2 + 4x +3y2 + 12y = 10 Factor out the GCF’s... How do you write an equation in standard form? Complete the square for each variable. 2(x2 + 2x ) + 3(y2 + 4y ) = 10 +1 +4 +2 +12 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 24 24 (x +1)2 + (y +2)2 = 1 12 8

  38. Example: Graph the hyperbola9x2 – 4y2 + 18x + 16y – 43 = 0 Group the x’s and y’s together... 9x2 + 18x –4y2 + 16y = 43 Factor out the GCF’s... 9(x2 + 2x ) – 4(y2 – 4y ) = 43 +1 +4 + 9 – 16 Let’s try another…with graphing. Complete the squares ... 9(x + 1)2 – 4(y – 2)2 =36 9(x + 1)2 – 4(y – 2)2 = 36 36 36 36 (x +1)2 – (y – 2)2 = 1 4 9 Center (–1, 2) a = 2b = 3

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