The eEquation of a Circle

Adaptif Hal.: 2 Isi dengan Judul Halaman Terkait The eEquation of a Circle

Adaptif Hal.: 3 Isi dengan Judul Halaman Terkait CIRCLE IS DEFINED AS SET OF POINTS THAT WITH THE SAME DISTANCE TOWARDS A PARTICULAR REFERENCE POINT, AND IT IS MENTIONED AS CIRCLE CENTRAL AND THE SAME DISTANCE CALLED RADIUS The eEquation of a Circle

Adaptif Hal.: 4 Isi dengan Judul Halaman Terkait o r The eEquation of a Circle

Adaptif Hal.: 5 Isi dengan Judul Halaman Terkait The eEquation of a Circle The equation of the circle with center of O(0,0) and radius r The equation of the circle with center of P(a,b) and radius r The eEquation of a Circle

Adaptif Hal.: 6 Isi dengan Judul Halaman Terkait o r T (x,y) OT = r x + y = r 222 ( x 2 - x 1 ) + ( y 2 - y 1 ) = r 22 ( x - 0 ) + ( y - 0 ) = r 22 X Y

Adaptif Hal.: 7 Isi dengan Judul Halaman Terkait The eEquation of a Circle

Adaptif Hal.: 8 Isi dengan Judul Halaman Terkait Determine the circle equation that center to point O (0,0) and : a. radius of 2 b. through the point (3,4) Exercise The eEquation of a Circle

Adaptif Hal.: 9 Isi dengan Judul Halaman Terkait P (a,b ) r T (x,y) PT = r (x- a) + (y-b) = r 2 22 ( x 2 - x 1 ) + ( y 2 - y 1 ) = r 22 ( x - a ) + ( y - b ) = r 22 O X Y

Adaptif Hal.: 10 Isi dengan Judul Halaman Terkait The eEquation of a Circle

Adaptif Hal.: 11 Isi dengan Judul Halaman Terkait Determine the circle equation if : a. Center to P (3,2) and radius of 4 b. Center to point Q (2,-1) and through the point of R(5,3) Exercise The eEquation of a Circle

Adaptif Hal.: 12 Isi dengan Judul Halaman Terkait

Adaptif Hal.: 13 Isi dengan Judul Halaman Terkait ELLIPSE

Adaptif Hal.: 14 Isi dengan Judul Halaman Terkait Ellipse Indicators 1. Explaining understanding of ellipse. 2. Determining ellipse terms. 3.Determining ellipse equation 4.Drawing graph of ellipse equation Base Competence: 3. Applying ellipse concept Standard Competence Applying cone section concept in solving problem.

Adaptif Hal.: 15 Isi dengan Judul Halaman Terkait Ellipse Indicators 1. Explaining understanding of ellipse. 2. Determining ellipse terms. 3. Determining ellipse equation. 4. Drawing graph of ellipse equation.

Adaptif Hal.: 16 Isi dengan Judul Halaman Terkait Ellipse Definition of Ellipse Ellipse is position place of points on the flat surface which has total distance towards certain two points that is constant.

Adaptif Hal.: 17 Isi dengan Judul Halaman Terkait See this ellipse picture Ellipse The terms in ellipse: 1.F 1 and F 2 called focus. If T is random point in ellipse then TF 1 + TF 2 = 2a, F 1 F 2 = 2c, and 2a > 2c. 2. A1A2 is long axis (mayor)= 2a. B1B2 is short axis (minor) = 2b, that’s why a > b. b B1B1 a T A2A2 E D A1A1 B2B2 (0,-b) (0,b) F1F1 F2F2 P (c, 0)(- c, 0) K L continue Ellipse terms

Adaptif Hal.: 18 Isi dengan Judul Halaman Terkait Ellipse 3. Lotus Rectum is line segment that limits ellipse, upright straight to mayor axis through focus (DE and KL), length of Lotus Rectum DE = KL = 4. Center point (P) is intersection point toward mayor axis with minor axis. 5. Top point of ellipse is point A 1, A 2, B 1, B 2.

Adaptif Hal.: 19 Isi dengan Judul Halaman Terkait Ellipse 1. Ellipse equation that center to O(0,0) Ellipse equation : TF 1 + TF2 = 2a + = 2a = 2a - Squaring left side and right so we get…… (a 2 - c 2 ) x 2 + a 2 y 2 = a 2 (a 2 -c 2 )... (i), If point T at top point in minor axis (0,b) then …. b 2 =a 2 – c (ii) Equation (ii) is substituted to equation (i) then we get: Ellipse Equation

Adaptif Hal.: 20 Isi dengan Judul Halaman Terkait Ellipse Example Determine ellipse equation with top point (13,0) and focus F1(-12, 0) and F2(12,0). Answer: Given ellipse center O(0,0) Top Point (13,0) a = 13 Focus point (-12,0) and (12,0) c = 12 Main axis is X, so the equation:

Adaptif Hal.: 21 Isi dengan Judul Halaman Terkait Ellipse 2.Ellipse equation that center to P (m,n) a. Ellipse equation with center point (m, n): b. Main axis y = n, with the length 2a and minor axis is x = n, with the length 2b. 3. Focus point F 1 (m-c, n) and F 2 ( m + c, n ) 4. Top point A(m-a, n) and B ( m + a, n ) 5.Length of lactus rectum (LR) = with O B C D P(m,n) X= m X Y A F1F1 F2F2 m

Adaptif Hal.: 22 Isi dengan Judul Halaman Terkait Ellipse Example: Determine the ellipse equation with focus F 1 (1,3) and F 2 (7,3) and the top (10,3). Focus (1,3) and (7,3) = m-c = 1, m + c = 7 with the elimination gotten m=4 and c= 3 Center P (m,n) = P (4,3) m = 3 Top(10,3) m + a= 10 a= 6 b 2 = a 2 –c 2 = = = 27 Main axis y=3, so ellipse equation become: Answer:

Adaptif Hal.: 23 Isi dengan Judul Halaman Terkait Ellipse General form of ellipse equation Ellipse equation has general form: Relation between equation and equation as follows: If A > B, then A = a 2, B = b 2, C=-2a 2 m, D= -2b 2 n, E= a 2 m 2 + b 2 n 2 - a 2 b 2 If A < B, then A = b 2, B = a 2, C=-2b 2 m, D= -2a 2 n, E= a 2 m 2 + b 2 n 2 - a 2 b 2

Adaptif Hal.: 24 Isi dengan Judul Halaman Terkait Ellipse Example: Find the top point and focus of ellipse that has equation 4x 2 + 9y 2 -16x+ 18y -11=0. Answer: Given ellipse equation: 4x 2 + 9y 2 -16x+ 18y -11=0. A=4, B= 9, C= -16, D=18, E= -11 b2 = A = 4 b = 2 A2 = B = 9 a = 3 C = -2 b 2 m D= -2a 2 mC 2 = a 2 –b 2 = 9 -4 = 5 -16= m18= nC = -16= -8m18= -18n 2= m-1 = n Center P(m,n) P(2, -1) FocusF2(m-c, n)=F2 and F2(m+c, n)=F2

Adaptif Hal.: 25 Isi dengan Judul Halaman Terkait Ellipse The equation of tangent line through the point (x 1, y 1 ) in ellipse 1. For ellipse equation equation of tangent line through (x1, y1) in ellipse is: 2. For ellipse equation tangent line equation through (x 1, y 1 ) in ellipse is:

Adaptif Hal.: 26 Isi dengan Judul Halaman Terkait Elips Persamaan garis singgung dengan gradien p Pada elips atau,adalah y= p Untuk elips dengan persamaan: Persamaan garis singgungnya adalah: y - n = p(x-m)

Adaptif Hal.: 27 Isi dengan Judul Halaman Terkait Ellipse Equation of tangent line with gradient p In ellipse or, is y= p For ellipse with the equation: The tangent line is: y - n = p(x-m)

Adaptif Hal.: 28 Isi dengan Judul Halaman Terkait Ellipse Example: Determine the tangent line equation of this ellipse. a. At point (4, 3) b. At point(5,-3) Answer: a.Given that : (4,3) x 1 = 4 and y 1 = 3 Equation of tangent line:

Adaptif Hal.: 29 Isi dengan Judul Halaman Terkait Ellipse b. Given that: center (m, n) = (1, -2) ( 5, -3) y 1 = -3 Equation of tangent line:

Adaptif Hal.: 30 Isi dengan Judul Halaman Terkait Ellipse

Adaptif Hal.: 31 Isi dengan Judul Halaman Terkait

Adaptif Hal.: 32 Isi dengan Judul Halaman Terkait

Adaptif Hal.: 33 Isi dengan Judul Halaman Terkait Parabola Parabola equation top 0(0,0) y 2 = 4px a. Top (0,0) b. Symmetry Axis = x c. Focus F(p,0) d. Directory x = -p (0,0) X d:X=-P F(P,0) Y

Adaptif Hal.: 34 Isi dengan Judul Halaman Terkait Parabola Parabola equation top in 0(0,0) and focus on F(-p,0) is Y 2 = -4px X Y (0,0)F(P,0) d:X=-P

Adaptif Hal.: 35 Isi dengan Judul Halaman Terkait Parabola Parabola equation top in 0(0,0) and focus on F(0,p) is x 2 = -4py X Y F(0,p) (0,0) d:y=-P

Adaptif Hal.: 36 Isi dengan Judul Halaman Terkait Parabola Parabola equation top in 0(0,0) and focus on F(0,-p) is x 2 = -4py X Y F(0,-p) (0,0) d: y=p

Adaptif Hal.: 37 Isi dengan Judul Halaman Terkait Parabola Example: 1. From next parabolas, find the focus coordinates, equation of symmetric axis, directory equation and the length of lactose rectum a. y 2 = 4x c. x 2 = -8y b. y 2 = -12x d. x 2 = 6y Answer: a.y 2 =4px y 2 = 4x, then p = 1 This parabola is horizontal parabola that right opened. (i) Coordinate of focus point F(p,0) F(1,0) (ii) Symmetric axis that close to axis x, then the equation y = 0 (iii) Directory equation : x = -p x = -1 (iv) The length of lactose rectum (LR)= 4p = 4.1=4

Adaptif Hal.: 38 Isi dengan Judul Halaman Terkait Parabola b. y 2 =-p4x y 2 = -12x, then 4p = 12 p = 3 This parabola is horizontal parabola that left opened (i) Coordinate of focus point F(-p,0) F(-3,0) (ii) Symmetric axis that close to axis X, then the equation of y = 0 (iii) Directory equation: x = -p x = 3 (iv) The length of lactose rectum (LR) = 4p = 4. 3= 12 c. x 2 = -p4y x 2 = -8y, then 4p = 8 p = 2 This parabola is horizontal parabola that below opened (i) Coordinate of focus point F(0,-p) F(0,-2) (ii) Symmetry axis that close to axis y, then the equation of X = 0 (iii) Directory equation: y = p y = 2 (iv) The length of lactose rectum (LR) = 4p = 4. 2 = 8 d. Exercise

Adaptif Hal.: 39 Isi dengan Judul Halaman Terkait Parabola Parabola equation in top P(a,b) (y – b) 2 = 4p(x – a) x O(0,0) F(p,0 ) y P(a,b) F p (a+p,b) a a. Top point P(a,b) b. Focus point F(a+p,b) c. Directory x = -p+a d. Symmetry axis y = b e.

Adaptif Hal.: 40 Isi dengan Judul Halaman Terkait Parabola Example: Given that parabola equation 3x – y 2 + 4y + 8= 0 Determine: a. Top point c. Directory b. Focus point d. Symmetry axis Answer: Change parabola equation into general equation: 3x – y 2 + 4y + 8= 0 y 2 - 4y = 3x + 8 y 2 - 4y + 4 = 3x (y – 2) 2 = 3x + 12 (y – 2) 2 = 3(x + 4) Gotten that parabola equation (y – 2) 2 = 3(x + 4) is The flat parabola that right opened.

Adaptif Hal.: 41 Isi dengan Judul Halaman Terkait Parabola From that equation we get: a. Top point P(-4,2) b. 4p = 3 then p = Focus point F(a+p,b) c. Directory equation : d. Symmetry axis : y = 2 x O(0,0) P(-4,2) F y

Adaptif Hal.: 42 Isi dengan Judul Halaman Terkait Parabola Exercise: a. Find the parabola equation that top in (2,4) and the focus (-3,4) b. Find the equation parabola that has focus point F(2-3) and the directory equation y = 5

Adaptif Hal.: 43 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola A.Equation of the parabola tangent line through point of A(x 1,y 1 ) yy 1 = 2p(x+x 1 ) x y A(x1,y1)

Adaptif Hal.: 44 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola Parabola Equation through point of A(x1,y1) and presented in this table Parabola EquationEquation of Tangent Line y 2 = 4pxyy 1 = 2p(x+x 1 ) y 2 = -4pxyy 1 = -2p(x+x 1 ) x 2 = 4pyxx 1 = 2p(y+y 1 ) x 2 = -4pyxx 1 = -2p(y+y 1 ) (y – b) 2 = 4p(x – a)(y-b)(y 1 -b)=2p(x+x 1 -2a) (y – b) 2 = -4p(x – a)(y-b)(y 1 -b)=-2p(x+x 1 -2a) (x– a) 2 = 4p(y – b)(x-a)(x 1 -a)=2p(y+y 1 -2b) (x– a) 2 = -4p(y – b)(x-a)(x 1 -a)=-2p(y+y 1 -2b)

Adaptif Hal.: 45 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola Example: 1.Determine the equation of tangent line of parabola y 2 = 8x at point (2,4) Answer : y 2 = 8x 4p = 8 p = 2 Point A(x1,y1) A(2,4) The equation of tangent line is yy 1 = 2p(x+x 1 ) y.4 = 2.2(x+2) 4y = 4(x+2) y = x+2

Adaptif Hal.: 46 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola 2. Determine the equation of parabola tangent line (x+1) 2 = -3(y-2) at point (2,-1) Answera : a = -1, b = 2, x 1 = 2 and y 1 = 1 (x+1) 2 = -3(y-2) -4p = -3 p = The equation of parabola tangent line at point A(2,-1) is (x - a)(x1 - a) = -2p(y + y1 - 2b) (x +1)(2 +1) = -2. (y - 1 – 2.2) (x + 1)(3) = 6(x + 1) = - 3(y – 5) 2(x + 1) = -(y – 5) 2x + 2 = -y + 5 y = -2x + 3

Adaptif Hal.: 47 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola B. The equation of parabola tangent line that has gradient m Parabola EquationEquation of tangent line y2 = 4pxy = mx + y2 =- 4px y = mx - x 2 = 4pyy = mx – m 2 p x 2 = -4pyy = mx + m 2 p (y – b) 2 = 4p(x – a)(y – b) = m(x – a) + (y – b) 2 = -4p(x – a)(y – b) = m(x – a) - (x– a) 2 = 4p(y – b)(y – b) = m(x – a) – m 2 p (x– a) 2 = -4p(y – b)(y – b) = m(x – a) + m 2 p

Adaptif Hal.: 48 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola Example: 1.Find the equation of parabola tangent line y 2 = 8x that has gradient 2 Answer: Parabola y 2 = 8x 4p = 8 p = 2 Then the equation of tangent line is: y = mx + y = 2x + 1

Adaptif Hal.: 49 Isi dengan Judul Halaman Terkait Equation of tangent line in parabola 2. Determine the equation of parabola tangent line (y + 5) 2 = -8(x – 2) that has gradient 3 Answer : Parabola (y + 5) 2 = -8(x – 2) -4x = -8 p = 2 Top of P(2,-5) So the equation of tangent line is y – b = m(x – a) – y + 5 = 3(x – 2) – 3y + 15 = 9(x – 2) -2 3y + 15 = 9x – 20 9x – 3y + 35 = 0 y = 3x -

Adaptif Hal.: 50 Isi dengan Judul Halaman Terkait Hyperbola A. Hyperbola is position of points on the flat surface which has distance difference towards two certain points is constant.. Two certain points is Focus (reach point). x y 0 Y = BAF(C,0)F’(-C,0) A. Equation of Center Hyperbola(0,0) N a. Center O(0,0) b. Focus F’(-C,0) and F(C,0) c. Top of A(-a,0) and B(a,0) d. Symmetry axis - Main axis of axis x - The flock axis is axis y e. Real axis AB = 2a f. Imaginer axis MN = 2b KM L E D g. Asymptote, y = +

Adaptif Hal.: 51 Isi dengan Judul Halaman Terkait Hyperbola x y 0 Y = B A F(0,C) F’(0,-C) B. Hyperbola Equation N a. Center O(0,0) b. Focus F’(0,-C) and F(0,C) c. Top of A(0,-a) and B(0,a) d. Symmetry axis - Main Axis of axis y - Flock axis is x e. Real axis AB = 2a f. Imaginer Axis MN = 2b K M L E D g. Asymptote, y = + or b 2 y 2 – a 2 x 2 = a 2 b 2

Adaptif Hal.: 52 Isi dengan Judul Halaman Terkait Hyperbola Example : 1.Find the hyperbola equation if the focus point is F’(-13,0) and F(13,0) while the top (-5,0) and (5,0) Answer : Center (0,0) a = 5, c = 13 b 2 = c 2 – a 2 = 13 2 – 5 2 = 169 – 25 = 144 The main axis of axis X, then the hyperbola equation is:

Adaptif Hal.: 53 Isi dengan Judul Halaman Terkait Hyperbola 2. Given that hyperbola equation of Answer : and Center (0,0) Top (-a,0)=(-4,0) and (a,0) = (4,0)

Adaptif Hal.: 54 Isi dengan Judul Halaman Terkait Hyperbola A. Hyperbola Equation at center P(m,n) x y 0 Y = BAF(C,0)F’(-C,0) N a. Center P(m,n) b. Focus F’(m-C,0) and F(m+C,0) c. Top of A(m-a,0) and B(m+a,0) d. Symmetry Axis - Main axis of axis y = n - Flock axis is y = m e. Real Axis AB = 2a f. Imaginer axis MN = 2b KM L E D g. Asymptote, y-n = + (x- a) P

Adaptif Hal.: 55 Isi dengan Judul Halaman Terkait Hiperbola Example: 1.Determine the equation of hyperbola if the focus point F’(-2,-3) and F(8,-3) and top point is (7,-3) Answer: focus F’(-2,-3) and F(8,-3) Distance from center to focus c = 8 – 3 = 5 Top (7,3) Distance from center with the top a = 7 – 3 = 4 b 2 = c 2 – a 2 = 5 2 – 4 2 = 25 – 16 = 9 So the equation of hyperbola is or 9(x-3) 2 – 16(y+3) 2 = 144 9x 2 – 16y 2 – 54x -96y – 207 = 0

Adaptif Hal.: 56 kkkkIsi dengan Judul Halaman Terkait Hyperbola 2. Determine center point, focus point, top point, length of lactus rectum and asymptote from Answer: Center point (4,-1)

Adaptif Hal.: 57 Isi dengan Judul Halaman Terkait Equation of Tangent Line in Hyperbola Equation of tangent line in hyperbola through T(x 1,y 1 ) Equation tangent line at point T(x 1,y 1 ) is at point T(x 1,y 1 ) is at point T(x 1,y 1 ) is at point T(x 1,y 1 ) is

Adaptif Hal.: 58 Isi dengan Judul Halaman Terkait Example 1 : Find the equation of tangent line in hyperbola At point (9, -4) EQUATION OF TANGENT LINE IN HYPERBOLA Answer: Equation of tangent line in hyperbola At point T(x 1,y 1 ) is So the equation of tangent line is : or x + 2y = 1

Adaptif Hal.: 59 Isi dengan Judul Halaman Terkait EQUATION OF TANGENT LINE IN HYPERBOLA Example 2 Determine the equation of tangent line in hyperbola At point (-4, -3) Answer : The equation of tangent line in hyperbola At point T(x 1,y 1 ) is So the equation of tangent line is : x = - 4

Adaptif Hal.: 60 Isi dengan Judul Halaman Terkait

Adaptif Hal.: 61 Isi dengan Judul Halaman Terkait