STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS WRITING AND GRAPHING EQUATIONS OF CONICS GRAPHS OF RATIONAL FUNCTIONS STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics. CIRCLE (x – h) 2 + (y – k) 2 = r 2 Horizontal axis Vertical axis PARABOLA (y – k) 2 = 4p (x – h) (x – h) 2 = 4p (y – k) ELLIPSE (x – h) 2 (y – k) 2 + = 1 a 2 b 2 HYPERBOLA (x – h) 2 (y – k) 2 – = 1 b 2 a 2 (y – k) 2 (x – h) 2 – = 1 b 2 Naserellid

Naserellid (–2, 1) Write an equation of the parabola 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). (–2, 1) SOLUTION Choose form: Begin by sketching the parabola. Because the parabola opens to the left, it has the form : where p < 0. (y – k) 2 = 4p(x – h) Find h and k: The vertex is at (–2, 1), so h = – 2 and k = 1. Naserellid

The standard form of the equation is: 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). (–2, 1) SOLUTION (–3, 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. The standard form of the equation is: (y – 1) 2 = – 4(x + 2). Naserellid

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

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

Naserellid Write an equation of the ellipse with 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). (3, 5) (3, –1) (3, 6) (3, –2) SOLUTION Plot the given points and make a rough sketch. (x – h) 2 (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) 6 + ( –2) 2 (h, k) = , = (3, 2) Naserellid

Naserellid Write an equation of the ellipse with 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). (3, 5) (3, –1) (3, 6) (3, –2) SOLUTION Find a: The value of a is the distance between the vertex and the center. a = (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 Naserellid

Naserellid The standard form is : 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). (3, 5) (3, –1) (3, 6) (3, –2) SOLUTION 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) 2 (y – 2) 2 Naserellid

Naserellid (x + 1) 2 SOLUTION 4 Writing an Equation of a Translated Ellipse Graph (y + 1) 2 – = 1. (x + 1) 2 4 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. (–1, –2) (–1, 0) (–1, –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. Naserellid

Naserellid (x + 1) 2 SOLUTION 4 (–1, 0) (–1, –1) (–1, –2) Writing an Equation of a Translated Ellipse Graph (y + 1) 2 – = 1. (x + 1) 2 4 SOLUTION (–1, –2) (–1, 0) (–1, –1) Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes. Naserellid

Naserellid The expression B 2 – 4AC is called the discriminate CLASSIFYING A CONIC FROM ITS EQUATION The equation of any conic can be written in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 which is called a general second-degree equation in x and y. The expression B 2 – 4AC is called the discriminate of the equation and can be used to determine which type of conic the equation represents. Naserellid

Naserellid CLASSIFYING A CONIC FROM ITS EQUATION CONCEPT SUMMARY CONIC TYPES The type of conic can be determined as follows: DISCRIMINATE (B 2 – 4AC) TYPE OF CONIC < 0, B = 0, and A = C Circle < 0, and either B  0, or A  C Ellipse = 0 Parabola > 0 Hyperbola If B = 0, each axis is horizontal or vertical. If B  0, the axes are neither horizontal nor vertical. Naserellid

the value of the discriminate is: Classifying a Conic Classify the conic 2 x 2 + y 2 – 4 x – 4 = 0. Classify the conic 4 x 2 – 9 y 2 + 32 x – 144 y – 5 48 = 0. SOLUTION SOLUTION Since A = 4, B = 0, and C = –9, the value of the discriminate is: Since A = 2, B = 0, and C = 1, the value of the discriminate is: B 2 – 4 AC = 0 2 – 4 (2) (1) = – 8 B 2 – 4 AC = 0 2 – 4 (4) (–9) = 144 Because B 2 – 4 AC < 0 and A  C, Because B 2 – 4 AC > 0,. the graph is an ellipse. , the graph is a hyperbola. Naserellid

WRITING AND GRAPHING EQUATIONS OF CONICS Naserellid