Warm-up: Complete the square: HW: pg803 (8 – 20 even, 30 – 50 even)

HW Answers: Pg. 795 – 796 (2 – 20 even, 26 – 34 even, 42 – 48 even)   2) 1 4) -1.4826 6) 3.8391 8) -0.6873 10) 63.4 12) 111.8 14) 63.4 16) 116.6 18) 141.3 20) 45 26) 45 28) 63.4 30) 77.5 32) 52.8 34) 58.7

10.2 THE PARAB LA Objective: Find the vertex, focus, and directrix of a parabola Graph a parabola

A conic section is a curve formed by the intersection of _________________________ a plane and a double cone.

Let's sort out this definition by looking at a graph: A parabola is a locus (collection) of all points P in the plane that are the same distance from a fixed point F as they are from a fixed line D. The point F is called the focus of the parabola, and the line D is its directrix. As a result, a parabola is the set of points P for which d(F, P) = d(P, D) Let's sort out this definition by looking at a graph: 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 Take a line segment perpendicular to the directrix and intersect with a line segment from the focus of the same length. This will be a point on the parabola and will be the same distance from each. directrix focus by symmetry we can get the other half

If the coefficient on x is negative the parabola opens to the left Based on this definition and using the distance formula we can get a formula for the equation of a parabola with a vertex at the origin that opens left or right. If the coefficient on x is negative the parabola opens to the left If the coefficient on x is positive the parabola opens to the right p is the distance from the vertex to the focus (or opposite way for directrix) 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 The equation for the parabola shown is: p p The parabola opens to the right and the vertex is 3 away from the focus.

Let's find the focus and directrix of the parabola: This is 4p Since the coefficient is negative, this parabola opens to the left. From the vertex count 4 in the negative direction to get the focus. The directrix is a line located the same distance from the vertex in the other direction. x = 4 p =4 p =4 focus (-4, 0)

We could make a line segment through the focus of the parabola parallel to the directrix with endpoints on the parabola. This segment is called the latus rectum. The length of the latus rectum is 4p. This is very helpful information when graphing a parabola because we then know how wide the parabola is. (-4, 8) latus rectum x = 4 p p The length of the latus rectum is 16 so it is 8 each way from the focus. focus 4p (-4, 0) (-4, -8)

Previously this school year you graphed parabolas that opened up or down. The only difference with the equation is the x and the y are in different places. Let's look at the steps to graph the parabola. What direction does this open? If the x is squared it opens up or down (depending on the sign of the coefficient). If the y is squared it is right or left. What is the length of the latus rectum? Add the directrix (not necessary for graphing but we want to see how it relates here). Draw the parabola containing these points. The length of the latus rectum is 4. Make a line segment 4 units long (2 each way) through the focus. What is p? The directrix is “p” away from the vertex in the opposite direction as the focus. - 4 = - 4p so p = 1. The focus is a away from the vertex in the direction the parabola opens. If the x is squared it opens up or down (depending on the sign of the coefficient). If the y is squared it is right or left. y = 1 (-2, -1) (0, -1) (2, -1)

Our parabola may have horizontal and/or vertical transformations Our parabola may have horizontal and/or vertical transformations. This would translate the vertex from the origin to some other place. The equations for these parabolas are the same but h is the horizontal shift and k the vertical shift: opens up opens down Vertex: (h, k) Focus: (h, k + p) Directrix: y = k – p opens right opens left Vertex: (h, k) Focus: (h + p, k) Directrix: x = h – p

Let's try one: Vertex? (h, k) so Vertex is (-1, 2) y is squared and 8 is positive so right. Opens? 4p = 8 so p = 2. Focus is 2 away from vertex in direction parabola opens. Focus? (1, 6) Length of latus rectum? x = -3 This is number in front of parenthesis which is 8, so 4 each way from focus. (-1, 2) (1, 2) Directrix? (1, -2) “p" away from the vertex so 2 away in opposite direction of focus.

The secret to doing these is NOT to memorize a bunch of formulas, but to DRAW A PICTURE. What if we were given the focus of a parabola was (-2, 2) and the vertex was (-5, 2). If we draw a picture we can figure out the equation and anything else we need to know. Just looking at this much graphed can you determine which way the parabola opens (and therefore what the standard form of the equation looks like) Focus is “p" away from vertex so p = 3 3 (-5, 2) (-2, 2) The focus must be inside the parabola so it must open to the right. simplified:

6 Ex) Find the vertex, focus, directrix and graph. Opens? Vertex? Right (y squared & no negative) opposites of these values Vertex? Focus? 4p = 6 so p = 3/2 Length of latus rectum? 6, so 3 each way x = -5/2 Since the focus was at (1/2, -1), to get the ends of the latus rectum, we'd need to increase the y value of the focus by 3 and then decrease the y value by 3. (look at the picture to determine this). (1/2, 2) (-1, -1) (1/2, -1) (1/2, -4) Directrix? 3/2 away from vertex

Find the standard form of the equation of the parabola given: the vertex is (2, -3) and focus is (2, -5) Because of the location of the vertex and focus this must be a vertical parabola that opens down Equation: (x – h)2 = 4p(y – k) |p| = 2 V Equation: (x – 2)2 = -8(y + 3) F The vertex is midway between the focus and directrix, so the directrix for this parabola is y = -1

Sneedlegrit: Find the standard form of the equation of the parabola given: the vertex is (-1, 2) and focus is (-1, 0) HW: pg803 (8 – 20 even, 30 – 50 even)