1. 11.5 Translation of Axes & the General Form

2. So far our conic sections in general form have looked like this: Ax 2 + Cy 2 + Dx + Ey + F = 0 But there can be another term, an xy term! Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 The xy term means the graph will have a major axis that is not parallel to the x- or y-axis. To decide the type of conic it is, follow these rules utilizing the discriminant: B 2 – 4AC If B 2 – 4AC < 0  circle or ellipse (circle if B = 0 & A = C) B 2 – 4AC = 0  parabola B 2 – 4AC > 0  hyperbola Ex 1) Identify the conic 3x 2 – 3xy – y 2 + 2x + 2y – 10 = 0 B 2 – 4AC = (–3) 2 – 4(3)(–1) = 9 + 12 = 21 > 0 hyperbola

3. Translation of Axes It is possible to talk about a graph of a conic from different vantage points on the coordinate grid. We will consider conics from a “new center” instead of just the origin. There are two types: (1) given an equation and a point  write a new equation in terms of this new vantage point (2) given an equation  find the center which is the vantage point x', read “x prime,” is a new x, the x has been moved (same goes for y) new point = old point – change x' = x – h andy' = y – k x = x' + hy = y' + k

4. Ex 2) Find the translation equation given the “new” origin. 5x 2 – 4y 2 + 10x – 16y + 29 = 0 ; (–1, 6) x' = x + 1y' = y – 6 x = x' – 1 y = y' + 6 every time we see x & y, we replace with new expression 5(x' – 1) 2 – 4(y' + 6) 2 + 10(x' – 1) – 16(y' + 6) + 29 = 0 5(x') 2 – 10x' + 5 – 4(y') 2 – 48y' – 144 + 10x' – 10 – 16y' – 96 + 29 = 0 5(x') 2 – 4(y') 2 – 64y' – 216 = 0

5. Ex 3) Translate the axes so that the graph is symmetric to at least one coordinate axis (basically the origin!). Find equation with respect to the new system and graph. *Complete the square to find the center* 9x 2 + y 2 + 36x – 8y + 43 = 0 9(x 2 + 4x + 4 ) + (y 2 – 8y + 16 ) = – 43 + 36 + 16 9(x + 2) 2 + (y – 4) 2 = 9 9 9 *center (–2, 4)x + 2 = x' y – 4 = y'

6. Homework #1107 Pg 568 #1, 3, 7, 9, 11, 13, 17, 21, 25, 27, 34, 36, 40, 42