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

Write an equation of the parabola whose vertex is at (–2, 1) 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) 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.

Write an equation of the parabola whose vertex is at (–2, 1) 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) 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).

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

(3, 2) SOLUTION r (– 1, – 2) (3, – 2) (7, – 2) (3, – 6) Graphing the Equation of a Translated Circle (3, 2) Graph (x – 3) 2 + (y + 2) 2 = 16. SOLUTION r The radius is r = 4 (– 1, – 2) (3, – 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.

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) 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)

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 = 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

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) 2 (y – 2) 2

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.

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.

The equation of any conic can be written in the form 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 discriminant of the equation and can be used to determine which type of conic the equation represents.

CLASSIFYING A CONIC FROM ITS EQUATION CONCEPT SUMMARY CONIC TYPES The type of conic can be determined as follows: DISCRIMINANT (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.

Classify the conic 2 x 2 + y 2 – 4 x – 4 = 0. Classifying a Conic Classify the conic 2 x 2 + y 2 – 4 x – 4 = 0. Help SOLUTION Since A = 2, B = 0, and C = 1, the value of the discriminant is: B 2 – 4 AC = 0 2 – 4 (2) (1) = – 8 Because B 2 – 4 AC < 0 and A  C, the graph is an ellipse.

Classify the conic 4 x 2 – 9 y 2 + 32 x – 144 y – 5 48 = 0. Classifying a Conic Classify the conic 4 x 2 – 9 y 2 + 32 x – 144 y – 5 48 = 0. Help SOLUTION Since A = 4, B = 0, and C = –9, the value of the discriminant is: B 2 – 4 AC = 0 2 – 4 (4) (–9) = 144 Because B 2 – 4 AC > 0, the graph is a hyperbola.

CLASSIFYING A CONIC FROM ITS EQUATION CONCEPT SUMMARY CONIC TYPES Back The type of conic can be determined as follows: DISCRIMINANT (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.

CLASSIFYING A CONIC FROM ITS EQUATION CONCEPT SUMMARY CONIC TYPES Back The type of conic can be determined as follows: DISCRIMINANT (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.