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Conic sections – curves that result from the intersection of a right circular cone and a plane. 9.1 Conic Sections. Circle Ellipse Parabola Hyperbola. 9.2 Quadratic Function: Any function of the form. y = a x 2 + b x + c
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Conic sections – curves that result from the intersection of a right circular cone and a plane. 9.1 Conic Sections Circle Ellipse Parabola Hyperbola
9.2 Quadratic Function: Any function of the form y = ax2 + bx + c a, b, c are real numbers & a 0 A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______ 1 4 3 x y -2 -1 -1 0 -3 0 0 3 -4 3 Formula for Vertex: X = -b 2a Plug x in to Find y The graph is a parabola. Where is the vertex? Where is the axis of symmetry?
Parabola Directrix Focus Focus Parabola Vertex Axis of Symmetry x Vertex Directrix 9.2 Parabola A Parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) that is not on the line. Vertex Form of a Parabola (opens up or down) = - + 2 x a ( y k ) h (opens right or left) • Vertex = (h, k) • Focus & Directrix = +/- 1 from vertex • 4a • Plotting Points of the parabola can be done by: • finding x/y intercepts and/or • choosing values symmetric to the vertex
= - + 2 x a ( y k ) h Step 1: Place equation in one of two vertex forms: Parabola Graphing (opens up or down) (opens right or left) Step 2: Find the vertex at : (h, k) & determine which direction the parabola opens Step 3: Use an x/y chart to plot points symmetric to the vertex & draw the parabola Step 4: Find the focus/directrix displacement with 1/(4a) Try the following Examples: #1) y2 = 4x #2) x2 = -16y #3) (y + 4)2 = 12(x + 2) #4) x2 – 2x – 4y + 9 = 0
(x – 2)2 + (y – (-3))2 = 32 h is 2. k is –3. r is 3. Standard circle equation(x – h)2 + (y – k)2 = r2 center = (h, k)radius = r 1.1 Circles Graph the circle: (x – 2)2 + (y + 3)2 = 9 Step 1: Find the center & radius Step 2: Graph the center point Step 3: Use the radius to find 4 points directly north, south, east, west of center. Center = (2, -3) Radius = 3
General Form of a Circle Equation(Changing from General to Standard Form) Standard circle equation : (x – h)2 + (y – k)2 = r2center = (h, k)radius = r General Form of a circle : x2 + y2 + Dx + Ey + F = 0 Change from General Form to Standard Form General Form: x2 +y2 –4x +6y +4 = 0 Move constant : x2 +y2 –4x +6y = -4 X’s & Y’s together : (x2 – 4x ) + (y2 + 6y ) = -4 Complete the Square: (x2 – 4x +4 ) + (y2 + 6y +9 )= -4 +4 +9 Factor (x – 2) (x –2) + (y + 3) (y + 3) = 9 Standard Form (x – 2)2 + (y + 3)2 = 9 Setup to Complete Square ½ (-4) = -2 ½ (6) = 3 (-2)2 = 4 (3)2 = 9
P P F1 F2 9.3 Ellipse Ellipse - the set of all points in a plane the sum of whose distances from two fixed points, is constant. These two fixed points are called the foci. The midpoint of the segment connecting the foci is the center of the ellipse. A circle is a special kind of ellipse. Since we know about circles, the equation of a circle can be used to help us understand the equation and graph of an ellipse.
(0, 5) (0, 3) (0, 0) (-4, 0) (4, 0) (0, -3) (0, -5) Comparing Ellipses to Circles circle:(x – h)2 + (y – k)2 = r2 Ellipse: (x – h)2 + (y – k)2 = 1 a2 b2 center = (h, k)center = (h, k) radius = rradiushorizontal = a radiusvertical = b We do not usually call ‘a’ and ‘b’ radii since it is not the same distance from the center all around the ellipse, but you may think of them as such for the purpose of locating vertices located at : (h + a, k) (h – a, k) (h, k + b) (h, k – b) Example1: Graph the ellipse: 25x2 + 16y2 = 400 Finding Ellipse Foci: (foci-radius)2 = (major-radius)2 – (minor-radius)2 Major and Minor axis is determined by knowing the longest and shortest horizontal or vertical distance across the ellipse. Example2: Calculate/Find the foci above.
Converting to Ellipse Standard Form Example: Graph the ellipse: 9x2 + 4y2 – 18x + 16y – 11 = 0 Move constant : 9x2 +4y2 –18x +16y = 11 X’s & Y’s together : (9x2 – 18x ) + (4y2 + 16y ) = 11 9(x2 – 2x ) + 4(y2 + 4y ) = 11 Complete the Square: 9(x2 – 2x + 1 ) + 4(y2 + 4y + 4) = 11 + 9+ 16 Factor 9(x – 1) (x –1) + 4(y + 2) (y + 2) = 36 9 (x – 1)2 + 4(x + 2)2 = 36 Standard Form 9(x – 1)2 + 4(y + 2)2 = 1 36 36 (x – 1)2 + (y + 2)2 = 1 4 9 Can you graph this?
Equation Center Major Axis Foci Vertices a2 > b2 and b2 = a2 – c2 (h, k) Parallel to the x-axis, horizontal (h – c, k) (h + c, k) (h – a, k) (h + a, k) b2 > a2 and a2 = b2 – c2 (h, k) Parallel to the y-axis, vertical (h, k – c) (h, k + c) (h, k – a) (h, k + a) y Major axis Focus (h + c, k) Vertex (h + a, k) Focus (h – c, k) Focus (h + c, k) (h, k) Major axis (h, k) Vertex (h – a, k) Vertex (h + a, k) x x Focus (h – c, k) Vertex (h + a, k) Standard Forms of Equations of Ellipses Centered at (h,k)
y Transverse axis Vertex Vertex x Focus Focus Center 9.4 Hyperbola A hyperbola is the set of points in a plane the difference whose distances from two fixed points (called foci) is constant y x Traverse Axis - line segment joining the vertices. When you graph a hyperbola you must first locate the center and direction of the traverse access -- parallel or horizontal Can you find the traverse axis, center, vertices and foci in the hyperbola above?
Graphing Hyperbola Hyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1 a2 b2 a2 b2 transverse axis: Horizontal transverse axis: Vertical center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2 Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h) center traverse-coordinate –a m = +/- y-displacement x-displacement Example1 : C Center: (3, 1) Traverse Axis: Horizontal Foci-displacement F2 = 4 + 25 = 29 F = 29 = 5.4
Graphing Hyperbola (Cont.) Hyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1 a2 b2 a2 b2 transverse axis: Horizontal transverse axis: Vertical center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2 Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h) center traverse-coordinate –a m = +/- y-displacement x-displacement Example2 : (y – 2)2– (x + 1)2= 1 9 16 C Center: (-1, 2) Traverse Axis: Vertical Foci-displacement F2 = 9 + 16 = 25 F = 25 = 5
Graphing Hyperbola (Completing the Square) Hyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1 a2 b2 a2 b2 transverse axis: Horizontal transverse axis: Vertical center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2 Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h) center traverse-coordinate –a m = +/- y-displacement x-displacement Example3 : 4x2 – y2 + 32x + 6y +39 = 0
Equation Center Transverse Axis Foci Vertices b2 = c2 – a2 (h, k) Parallel to the x-axis, horizontal (h – c, k) (h + c, k) (h – a, k) (h + a, k) b2 = c2 – a2 (h, k) Parallel to the y-axis, vertical (h, k – c) (h, k + c) (h, k – a) (h, k + a) y Vertex (h + a, k) Focus (h – c, k) Focus (h + c, k) (h, k) Focus (h + c, k) (h, k) Vertex (h + a, k) Vertex (h – a, k) Vertex (h + a, k) x x Focus (h – c, k) Standard Forms of Equations of Hyperbolas Centered at (h,k)