Mashadi Jurusan matematika FMIPA UNRI

Mashadi Jurusan matematika FMIPA UNRI
Pertemuan ke 2 dan ke 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graphing Lines slope and y-intercept 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

y = 2x + 1 y = 2x + 1 Run 1 y-intercept Rise b = 1 2 slope m = 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

y = -3x + 2 y = -3x + 2 Run 1 y-intercept Rise b = 2 -3 slope m = -3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

Run 3 y-intercept b = -1 Rise 2 slope 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

Run 2 y-intercept Rise b = 3 -1 slope 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

Graphing Lines x- and y- intercepts 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

2x + 3y = 5 find the x value when y = 0 x-intercept 2x + 3(0) = 5 2x + 0 = 5 2x = 5 find the y value when x = 0 y-intercept 2(0) + 3y = 5 0 + 3y = 5 3y = 5 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

4x - y = 6 find the x value when y = 0 x-intercept 4x - (0) = 6 4x - 0 = 6 4x = 6 find the y value when x = 0 y-intercept 4(0) - y = 6 0 - y = 6 -y = 6 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

4x + 2y = 3 find the x value when y = 0 x-intercept 4x + 2(0) = 3 4x + 0 = 3 4x = 3 find the y value when x = 0 y-intercept 4(0) + 2y = 3 0 + 2y = 3 2y = 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

3x - 2y = 7 find the x value when y = 0 x-intercept 3x - 2(0) = 7 3x - 0 = 7 3x = 7 find the y value when x = 0 y-intercept 3(0) - 2y = 7 0 - 2y = 7 -2y = 7 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Jeff Bivin -- LZHS

Y y=x y=x-1 y=x+4 y=x+k y=x-3 x=k O X 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Y y=-x y=-x+5 y=-x+2 y=-x-1 O -1 2 -c 5 X y=-x-c LALU MINTA SISWA MEMBUAT KESIMPULAN BAGAIMANA KALAU SISWA MEMBUAT KESIMPULAN 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

MASHADI, JURUSAN MATEMAIKA FMIPA UNRI
4/5/2019 Y y=2x y=2(x+3) y=2(x-1) y=2(x-4) y=2(x+k) O k X BIMBING SISWA MEMBUAT KESIMPULAN 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI MASHADI JURUSAN MATEMATIKA FMIPA UNRI

(x – a)2 + (y - b)2 = r2 Pusat lingkaran (a,b) , r = jari-jari
Persamaan Lingkaran Pusat (a,b) dan jari-jari r y (a, b) b x a (0,0) (x – a)2 + (y - b)2 = r2 Pusat lingkaran (a,b) , r = jari-jari 4/5/2019 Prof. Dr. Mashadi, Jurusan Matematika FMIPA Universitas Riau

Bentuk normal (Normal form) garis lurus
Y l P(x,y) o X n Q(cos , sin ) N Sumbu normal Vektor normal n yang merupakan unit vektor Apa yang sudah diketahui, (Titik N, , . . .) Persoalan kita adalah menentukan : Vektor normal n, Bentuk normalnya 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Misalkan P(x,y) sebarang titik pada garis l dan  sudut antara vektor OP =[x,y] dan n = [cos , sin ] Maka diperoleh x cos  + y sin  = p Yang disebut dengan bentuk normal (normal form) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Selanjutnya bagaimana bentuk ax + by + c = 0 kita ubah dalam bentuk normal Sehingga diperoleh : Ingat : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

contoh Tulislah persamaan 3x – 4y +10 = 0 dalam bentuk normal dan gambarlah grafiknya Penyelesaian : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Y N X o 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Penyelesaian 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Gambarlah garis yang persamaan parameternya adalah x = -3 – 2t dan y =1 + 3t  (-5,4) 3 (-3,1) -2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Capek, iseng dulu BERAPA JUMLAH KAKI GAJAH INI ?
2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Skip Intro CONIC SECTIONS 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI End Next

CONIC SECTIONS A conic section is the intersection of plane and a cone. Five cases involving conics sectioned at different angle circle ellipse parabola hyperbola triangle LOOK INTO THE GEOMETRICAL CONTRUCTION OF ELLIPSE, PARABOLA, AND HYPERBOLA. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conic Sections (1) Circle A circle is formed when i.e. when the plane  is perpendicular to the axis of the cones. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conic Sections (2) Ellipse An ellipse is formed when i.e. when the plane  cuts only one of the cones, but is neither perpendicular to the axis nor parallel to the a generator. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conic Sections (3) Parabola A parabola is formed when i.e. when the plane  is parallel to a generator. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conic Sections (4) Hyperbola A hyperbola is formed when i.e. when the plane  cuts both the cones, but does not pass through the common vertex. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Find an equation of the conis that has Focus is at (1,-2), directrik is the line 2x – y = 0; e = 1 Since e = 1, we know that the conic is parabola D P(x,y) PF = PD . F(1,-2) Then : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Since PF = PD x2 + 4y2 -4xy – 10x + 20y + 25 = 0 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Find an equation of the conis that has Focus is at (-5,-1), directrik is the line 2x +3y + 6 = 0; e = 1/2 Solution : Clearly ???? 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Parabola 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Here are some other applications of the parabola...

Definitions Parabola: d(F, P) = d(P, D) “The collection of points, P in the plane whose distance from a line, the directrix, and whose distance from a point, the focus, is equal. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Parabola The set of all points that are equidistant from a given point (focus) and a given line (directrix). focus directrix 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Parabola A parabola is the locus of a variable point on a plane so that its distance from a fixed point (the focus) is equal to its distance from a fixed line (the directrix x = - a). focus F(a,0) P(x,y) M(-a,0) x y O 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Form the definition of parabola, PF = PN standard equation of a parabola 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

axis of symmetry vertex Length of the latus rectum latus rectum (LL’) is 4a mid-point of FM = the origin (O) = vertex ??? Koordinat of L and L’ ???? L(a,2a) and L’(a,-2a) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Other forms of Parabola y2 = -4ax 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Other forms of Parabola x2 = 4ay 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Other forms of Parabola x2 = -4ay 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Since equations Ax2 + Ey=0 and Cy2 + Dx = 0 are eqivalent to 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

On the other hand x2 = 4ay or x2 = -4ay and y2 = 4ax or y2 = -4ax Then 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Find the simple (standart) equation of the parabola that has its focus at (0,-2) Direktrix y=2 Solution . F(0,-2) (-4,-2) (4,-2) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Vertex (0,0) Then a = 2 x2 = -8y 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Determine the focus and directrix of the parabola –3y2 – 12x = 0 :
Since y is squared, the parabola goes left or right… Solve for y y2 = 4px –3y2 = 12x y2 = –4x Solve for p 4p = –4 p = –1 Focus: (p, 0) Directrix: x = –p Focus: (–1, 0) Directrix: x = 1 Let’s see what this parabola looks like... 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Discuss the equations 4x2 – 25y = 0 and sketch the curve. Solutions First we write the equations x2 = (25/4)y Then 4a = 25/4 a = 25/16 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Vertex is at the origin Focus ? F(0,25/16) Latus rectum ? F(0,25/16)  (-25/8, 25/16)  (25/8, 25/16) and (-25/8, 25/16) (25/8, 25/16) Directrix ? Y = -25/16 Y=-25/16 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Discuss the equations x2 + 12y = 0 and sketch the curve. Solutions Then 4a = 12 a = 3 Vertex is at the origin Focus ? F(0,-3) Why ???? The Parabola opens upward or downward ? The Parabola opens downward Latus rectum ? 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Latus rectum (6,-3) and (-6,-3) Y=3  (-6,-3) F(0,-3)  (6,-3) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Standard Form of the Parabola
(y – k)2 = 4p(x – h) (x – h)2 = 4p(y – k) Vertex (h, k) Focus (h + p, k) Directrix (h – p, k) Axis of symmetri y=k Vertex (h, k) Focus (h, k + p) Directrix (h, k – p) Axis of symmetri x=h 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

L L Find 1. Standart form of the parabola ? 2. Coordinat of the focus 3. Equations of the directrix 4. Axis of the symetri 5. Coordinat of the L and L’ 5. Length of the Latus Rectum 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI Copyright © Oswego City School District Regents Exam Prep Center

Find 1. Standart form of the parabola ? 2. Coordinat of the focus 3. Equations of the directrix 4. Axis of the symetri 5. Coordinat of the L and L’ 5. Length of the Latus Rectum 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1). SOLUTION (–2, 1) Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is p = (–3 – (–2)) 2 + (1 – 1) 2 = 1 so p = 1 or p = – 1. Since p < 0, p = – 1. (–3, 1) The standard form of the equation is (y – 1) 2 = – 4(x + 2). 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graph the following parabola
x2 - 2x + 8y + 9 = 0 (x2 - 2x ) = -8y - 9 (x2 - 2x ) = -8y – 9 (x - 1)2 = -8y - 8 opens (x - 1)2 = -8(y +1) 4p = 8 down a =2 or p = 2 Vertex: (1, -1) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

x2 - 2x + 8y + 9 = 0 p = 2 Vertex: (1, -1) Axis of symmetry: x = 1 Focus: (1 , -3) (1, -1) (-3, -3) (5, -3) Directrix: y = 1 (1, -3) Length of LR: 4p = 4(2) = 8 x = 1 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

y2 + x + y = 0 (y2 + y ) = -x (y2 + y ) = -x (y + )2 = -x opens (y + )2 = -(x - ) 4p = 1 left p = 1/4 Vertex: 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

y2 + x + y = 0 p = 1/4 Vertex: Axis of symmetry: y = -1/2 Focus: (0, -1/2) (0, 0) y = -1/2 Directrix: x = 1/2 (1/4 , -1/2) (0 , -1/2) Length of LR: 4p = 4(1/4) = 1 (0, -1) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Find the standard form of the equation, the vertex, focus, and directrix. The graph is oriented horizontally because the equation has a y2. Vertex (h, k) = (-1, 2) Focus (h + p, k) = (-1+1, 2) = (0, 2) Directrix (h – p, k) = (-2, 2) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse © Jill Britton, September 25, 2003 Statuary Hall in the U.S. Capital building is elliptic. It was in this room that John Quincy Adams, while a member of the House of Representatives, discovered this acoustical phenomenon. He situated his desk at a focal point of the elliptical ceiling, easily eavesdropping on the private conversations of other House members located near the other focal point. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipses An ellipse is the locus of a variable point on a plane so that the sum of its distance from two fixed points is a constant. P’(x,y) Alternative contructions for ellipses formula P’’(x,y) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Let PF1+PF2 = 2a where a > 0 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

standard equation of an ellipse 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

C1 major axis = 2a C1 vertex C2 loctus rectum C4 minor axis = 2b length of semi-major axis = a length of the semi-minor axis = b length of lactus rectum = Dari mana ??? 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Then length of lactus rectum = 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Other form of Ellipse 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Given an ellips wits its focus at (4,0), e = ½ and the center at the origin, find an equations of the ellipse Solutions (0,43) OF= c = 4 Since e = c/a = ½  a = 8 O (-8,0) (-4,0) F(4,0) (8,0) b2 = a2 – c2 = 82 – 42 = 48 = 43 (0,-43) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse Center (h, k) Definition of major and minor axis minor axis
major axis 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse major axis Center (h, k) minor axis 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse c2 = a2 - b2 Distance from center to vertex = a
Length of major axis = 2a Distance from center to co-vertex = b Length of minor axis = 2b c2 = a2 - b2 Distance from center to foci = c a a a c b c a b 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse The standard form of the ellipse with a center at (h,k) and a horizontal axis is…… 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse The ellipse with a center at (h,k) and a horizontal axis has the following characteristics…… Vertices (h a , k) Co-Vertices (h, k b) Foci (h c , k) ©Sellers, James 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse The standard form of the ellipse with a center at (h,k) and a vertical axis is…… 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Ellipse The ellipse with a center at (h,k) and a vertical axis has the following characteristics…… Vertices (h, k a) Co-Vertices (h b , k) Foci (h, k c) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI © Joan Bookbinder

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, 5) (3, –1) (3, 6) (3, –2) Plot the given points and make a rough sketch. (x – h) (y – k) 2 = 1 a 2 b 2 The ellipse has a vertical major axis, so its equation is of the form: Find the center: The center is halfway between the vertices. (3 + 3) ( –2) 2 (h, k) = , = (3, 2) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

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, 5) (3, –1) (3, 6) (3, –2) Find a: The value of a is the distance between the vertex and the center. a = (3 – 3) 2 + (6 – 2) 2 = = 4 Find c: The value of c is the distance between the focus and the center. c = (3 – 3) 2 + (5 – 1) 2 = = 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

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, 5) (3, –1) (3, 6) (3, –2) Find b: Substitute the values of a and c into the equation b 2 = a 2 – c 2 . b 2 = 4 2 – 3 2 b 2 = 7 b = 7 16 7 = 1 The standard form is (x – 3) (y – 2) 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graph the following Ellipse
Center: (2, -3) a = 5 in x direction b = 3 in y direction (2, 0) 3 (2, -3) (-3, -3) (7, -3) 5 5 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (2, -6)

a = 5 b = 3 c2 = a2 - b2 c2 = c2 = foci c2 = 16 c = 4 (2, 0) 3 (-2, -3) (2, -3) (6, -3) (-3, -3) (7, -3) 5 4 4 5 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (2, -6)

a = 5 b = c = 4 Find the Length of the LR. (2, 0) 3 (-2, -3) (2, -3) (6, -3) (-3, -3) (7, -3) 5 5 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (2, -6)

4x2 + 8x + 9y2 + 54y + 52 = 3 (4x2 + 8x ) + (9y2 + 54y ) = 4(x2 + 2x ) + 9(y2 + 6y + ) = 4(x + 1)2 + 9(y + 3)2 = 36 36 9 4 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Center: (-1, -3) a = 3 in x direction b = 2 in y direction (-1, -1) 2 (-1, -3) (-4, -3) (2, -3) 3 3 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (-1, -5)

a = 3 b = 2 c2 = a2 - b2 c2 = c2 = 9 - 4 foci c2 = 5 (-1, -1) 2 (-1, -3) (-4, -3) (2, -3) 3 3 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (-1, -5)

a = 3 b = c = Find the Length of the LR. (-1, -1) 2 (-1, -3) (-4, -3) (2, -3) 3 3 2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI (-1, -5)

9x2 + 36x + 4y2 - 40y = 88 (9x2 + 36x ) + (4y2 - 40y ) = 9(x2 + 4x + ) + 4(y2 - 10y ) = 9(x + 2)2 + 4(y - 5)2 = 324 324 36 81 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Center: (-2, 5) a = 9 in y direction b = 6 in x direction (-2, 14) 9 (-2, 5) (-8, 5) (4, 5) 6 6 9 (-2, -4) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

a = 9 b = 6 c2 = a2 - b2 c2 = c2 = (-2, 14) c2 = 45 foci 9 (-2, 5) (-8, 5) (6, 5) 6 6 9 (-2, -4) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Hyperbolas 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Definition The conic section formed by a plane which intersects both of the right conical surfaces Formed when or when the plane is parallel to the axis of the cone 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Definition A hyperbola is the set of all points in the plane where the difference between the distances from two fixed points (foci) is a constant Experiment with definition 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Let PF2 -PF1 = 2a where a > 0 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

standard equation of a hyperbola 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Sumbu transpersal vertex lotus rectum Sumbu konjugate length of lactus rectum = length of the semi-transverse axis = a length of the semi-conjugate axis = b 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

asymptote Dari mana ??? equation of asymptote : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Other form of Hyperbola : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Standard Equation 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 1 Find the equation of the hyperbola with foci at (7, 1) and (-3 ,1) whose transverse axis is 8 units long. Does the hyperbola open up and down or right and left? Find the center: Use the midpoint formula 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 1 Find the equation of the hyperbola with foci at (7, 1) and (-3 ,1) whose transverse axis is 8 units long. Now I’ve got (h, k) and all I need is a and b. a is the distance from the center to the vertex c is the distance from the center to the foci b2 = c2 - a2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 1 Find the equation of the hyperbola with foci at (7, 1) and (-3 ,1) whose transverse axis is 8 units long. The transverse axis is 8 units long, so the distance from the center to a vertex is 4. (a=4) The distance from the center (2, 1) to a focus (7,1) = 5 (c=5) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 1 Find the equation of the hyperbola with foci at (7, 1) and (-3 ,1) whose transverse axis is 8 units long. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 2 Find the coordinates of the center, foci and vertices of the graph of Does the hyperbola open up and down or right and left? Find the center: (2, -4) Vertices are: (h, ka) (2, -10) and (2, 2) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 2 Find the coordinates of the center, foci and vertices of the graph of The foci are (h, kc) and a2 + b2 = c2 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 3 (the hard one) Find the coordinates of the center, foci and vertices of the graph of: Rearrange and group Factor GCF for each variable Complete the square (you try it) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 3 (the hard one) Find the coordinates of the center, foci and vertices of the graph of: 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Example 3 (the hard one) Find the coordinates of the center, foci and vertices of the graph of: Find a, b & c Center = Foci = Vertices = 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graphing the Equation of a Translated Hyperbola Graph (y + 1) 2 – = 1. (x + 1) 2 4 (–1, –2) (–1, 0) (–1, –1) SOLUTION The y 2-term is positive, so the transverse axis is vertical. Since a 2 = 1 and b 2 = 4, you know that a = 1 and b = 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 2a = 2 units high and 2b = 4 units wide. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graphing the Equation of a Translated Hyperbola Graph (y + 1) 2 – = 1. (x + 1) 2 4 (–1, –2) (–1, 0) (–1, –1) SOLUTION The y 2-term is positive, so the transverse axis is vertical. Since a 2 = 1 and b 2 = 4, you know that a = 1 and b = 2. Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

How do you write an equation in standard form?
Example: Write the equation of the ellipse in standard form x2 + 3y2 + 4x + 12y – 10 = 0 Group the x’s and y’s together... 2x2 + 4x y2 + 12y = 10 Factor out the GCF’s... 2(x2 + 2x ) + 3(y2 + 4y ) = 10 +2 +12 Complete the square for each 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 (x +1)2 + (y +2)2 = 1 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Let’s try another…with graphing.
Example: Graph the hyperbola 9x2 – 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 + 9 – 16 Complete the squares ... 9(x + 1)2 – 4(y – 2)2 =36 9(x + 1)2 – 4(y – 2)2 = 36 (x +1)2 – (y – 2)2 = 1 Center (–1, 2) a = b = 3 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Hyperbola Center (h, k) c2 = a2 + b2 Slope of Asymptotes
Distance from center to vertex = a Distance from center to co-vertex = b Distance from center to foci = c Center (h, k) Slope of Asymptotes a c c2 = a2 + b2 b Length of transverse axis = 2a Length of conjugate axis = 2b 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Graph the following Hyperbola
Center: (-1, 5) a = 4 in x direction b = 7 in y direction (-1, 12) 7 (-1, 5) (-5, 5) 4 4 (3, 5) 7 2019/4/5 (-1, -2) Mashadi Jurusan matematika FMIPA UNRI

Center: (-1, 5) a = b = 7 a2 + b2 = c2 (-1, 12) = c2 = c2 65 = c2 7 foci (-1, 5) (-5, 5) 4 4 (3, 5) 7 2019/4/5 (-1, -2) Mashadi Jurusan matematika FMIPA UNRI

Asymptotes (-1, 12) 7 (-1, 5) (-5, 5) 4 4 (3, 5) 7 2019/4/5 (-1, -2) Mashadi Jurusan matematika FMIPA UNRI

Center: (-2, 3) a = 6 in y direction b = 3 in x direction (-2, 9) 6 (-2, 3) (-5, 3) 3 3 (1, 3) 6 2019/4/5 (-2, -3) Mashadi Jurusan matematika FMIPA UNRI

Center: (-2, 3) a = b = 3 a2 + b2 = c2 (-2, 9) = c2 = c2 45 = c2 6 foci (-2, 3) (-5, 3) 3 3 (1, 3) 6 2019/4/5 (-2, -3) Mashadi Jurusan matematika FMIPA UNRI

Asymptotes (-2, 9) 6 (-2, 3) (-5, 3) 3 3 (1, 3) 6 2019/4/5 (-2, -3) Mashadi Jurusan matematika FMIPA UNRI

a = 6 b = 3 Find the Length of the LR. (-2, 9) 6 (-2, 3) (-5, 3) 3 3 (1, 3) 6 (-2, -3) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Center: (-2, 3) Vertices: (-2, 9) (-2, -3) Co-Vertices: (-5, 3) (1, 3) Foci: (-2, 9) Length of Transverse axis: 12 Length of Conjugate axis: 6 6 Asymptotes: (-2, 3) (-5, 3) 3 3 (1, 3) Length of LR: 3 6 Endpoints of LR: 2019/4/5 (-2, -3) Mashadi Jurusan matematika FMIPA UNRI

4x2 + 8x - 9y2 + 54y - 53 = 168 (4x2 + 8x ) - (9y2 - 54y ) = 4(x2 + 2x ) - 9(y2 - 6y ) = 4(x + 1)2 - 9(y - 3)2 = 144 144 36 16 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Center: (-1, 3) a = 6 in x direction (-1, 7) b = 4 in y direction 4 (-1, 3) 6 6 (-7, 3) (3, 5) 4 2019/4/5 (-1, -1) Mashadi Jurusan matematika FMIPA UNRI

Center: (-1, 3) a = b = 3 foci a2 + b2 = c2 (-1, 7) = c2 = c2 4 52 = c2 (-1, 3) 6 6 (-7, 3) (5, 3) 4 2019/4/5 (-1, -1) Mashadi Jurusan matematika FMIPA UNRI

Asymptotes (-1, 7) 4 (-1, 3) 6 6 (-7, 3) (5, 3) 4 2019/4/5 (-1, -1) Mashadi Jurusan matematika FMIPA UNRI

Center: (-1, 3) Vertices: (-7, 3) (5, 3) Co-Vertices: (-1, 7) (-1, -1) Foci: (-1, 7) Length of Transverse axis: 12 4 Length of Conjugate axis: 8 (-1, 3) Asymptotes 6 6 (-7, 3) (5, 3) 4 2019/4/5 (-1, -1) Mashadi Jurusan matematika FMIPA UNRI

Rectangular Hyperbola If b = a, then The hyperbola is said to be rectangular hyperbola. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

equation of asymptote : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Problem solving (Bonus)
Find the coordinates of the center, foci and vertices of the graph of Find the focus, length of lotus rectum, if graph rotated through 450 about the origin 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

If the rectangular hyperbola x2 – y2 = a2 is rotated through 45o about the origin, then the coordinate axes will become the asymptotes. equation becomes : 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Simple Parametric Equations and Locus Problems x = f(t) y = g(t) parametric equations parameter Combine the two parametric equations into one equation which is independent of t. Then sketch the locus of the equation. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Equation of Tangents to Conics general equation of conics : Steps : Differentiate the implicit equation to find Put the given contact point (x1,y1) into to find out the slope of tangent at that point. (3) Find the equation of the tangent at that point. 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

OR 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Definition PF = PN PF1 + PF2 = 2a | PF1 - PF2 | = 2a 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Standard Equation 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Directrix x = -a 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Vertices (0,0) A1(a,0), A2(-a,0), B1(0,b), B2(0,-b) A1(a,0), A2(-a,0) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Axes axis of parabola = the x-axis major axis = A1A2 minor axis =B1B2 transverse axis =A1A2 conjugate axis =B1B2 where B1(0,b), B2(0,-b) 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Length of lantus rectum LL’ 4a 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Asymptotes ---- 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Conics Parabola Ellipse Hyperbola Graph Parametric representation of P 2019/4/5 Mashadi Jurusan matematika FMIPA UNRI

Mashadi Jurusan matematika FMIPA UNRI

Similar presentations

Presentation on theme: "Mashadi Jurusan matematika FMIPA UNRI"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Mashadi Jurusan matematika FMIPA UNRI

Similar presentations

Presentation on theme: "Mashadi Jurusan matematika FMIPA UNRI"— Presentation transcript:

Similar presentations

About project

Feedback