Intro to Conic Sections

It all depends on how you slice it! Start with a cone:

IF we slice the cone, parallel to the base, what do we get? A Circle!

IF we slice the cone at an angle, what do we get now? An Ellipse!

IF we just take a slice from the lateral face of the cone, what do we get? A Parabola!

Finally, let’s take a slice from the lateral face, perpendicular to the base A Hyperbola!

These shapes are called Conic Sections These shapes are called Conic Sections. We can use Algebra to describe the equations and graphs of these shapes.

10.2 - Parabolas

Review: No Calcs! Find the distance between the given points. 1. (2, 3) and (4, 1) 2. (4, 6) and (3, –2) 3. (–1, 5) and (2, –3) HMMM. . . Remember something called the Distance Formula?  

Solutions      

Let’s review Parts of a parabola The Vertex Form of a Parabola looks like this: y = a(x-h)² + k ‘a’ describes how wide or narrow the parabola will be. These are the roots Roots are also called: -zeros -solutions - x-intercepts This is the y-intercept, It is where the parabola crosses the y-axis This is the vertex, V (h, k) This is the called the axis of symmetry, a.o.s. Here a.o.s. is the line x = 2

New Vocabulary The focus  A point in the arc of the parabola such that all points on the parabola are equal distance away from the focus and the directrix A parabola is a set of all points that are the same distance form a fixed line and a fixed point not on the line The fixed point is called the focus of the parabola. The fixed line is called the directrix. The line segment through a focus of a parabola, perpendicular to the major axis , which has both endpoints on the curve is called the Latus Rectum. C is the distance between the focus and the vertex

Example Write an equation for a parabola with a vertex at the origin and a focus at (0, –7). Step 1: Determine the orientation of the parabola. Does it point up or down? Make a sketch. Since the focus is located below the vertex, the parabola must open downward. Use y = ax2. Step 2: Find a.

To find a, we use the formula: Write an equation for a parabola with a vertex at the origin and a focus at (0, –7). To find a, we use the formula: | a | = Note: c is the distance from the vertex to the focus. = Since the focus is a distance of 7 units from the vertex, c = 7. = 1 4(7) 28 4c Since the parabola opens downward, a is negative. So a = – . 1 28 An equation for the parabola is y = – x2. 1 28

WRITING EQUATIONS GIVEN … Write the EQ given Vertex(-2,4) and Focus(-2,2) Draw a graph with given info Use given info to get measurements C = distance from Vertex to Focus, so c = 2 (4-2 of the vertical coordinates) Also, parabola will open down because the focus is below the vertex Use standard form Y = a(x – h)² + k Need values for h,k, and a (h , k) = (-2 , 4) To find a, use formula a = 1/4c Therefore a = 1/8 Plug into formula Y = -1/8 (x + 2)² + 4 V(-2,4) C = 2 F(-2,2)

WRITING EQUATIONS GIVEN … Write the EQ given F(3,2) and directrix is x = -5 Draw a graph with given info Use given info to get measurements Vertex is in middle of directrix and focus. The distance from the directrix to the focus is 8 units. That means V = (-1 , 2) C = distance from V to F, so c = 4 Also, parabola will open right Use standard form x = a(y – k)² + h Need values for h,k, and a (h , k) = (-1 , 2) To find a, use formula a = 1/4c Therefore a = 1/16 Plug into formula x = 1/16 (y – 2)² – 1 X = -5 C = 4 F(3,2) Distance = 8 Vertex is in middle of directrix and F So V = (-1 , 2)

Let’s try one A parabolic mirror has a focus that is located 4 in. from the vertex of the mirror. Write an equation of the parabola that models the cross section of the mirror. The distance from the vertex to the focus is 4 in., so c = 4. Find the value of a. a = 1 4c = 1 4(4) = Since the parabola opens upward, a is positive. 1 16 The equation of the parabola is y = x2. 1 16

Let’s try one Write an equation for a graph that is the set of all points in the plane that are equidistant from point F(0, 1) and the line y = –1.

Let’s try one Write an equation for a graph that is the set of all points in the plane that are equidistant from point F(0, 1) and the line y = –1. An equation for a graph that is the set of all points in the plane that are equidistant from the point F (0, 1) and the line y = –1 is y = x2. 1 4

Example: Given the equation of a parabola, GRAPH and label all parts Start with the vertex V = (h,k) = (3,-1) Find the focus. C = 1/4a Since a = 1/8, then C = 2 Focus is at (3, -1 + 2)  (3,1) Label the directrix and A.O.S. Directrix is at y = -3 (the y coordinate of the vertex, minus the value of c) A.O.S.  x = 3 Find the latus rectum The length of the l.r. is | 1/a | Since a = 1/8, the l.r. =8 Plot and label everything X = 3 F(3,1) Y = -3 V(3,-1)

GRAPH and label all parts Start with the vertex V = (h,k) = (2,-1) Find the focus. C = 1/4a Since a = ¼ , then C = 1 Focus is at (2+1, -1)  (3,-1) Label the directrix and A.O.S. Directrix is at x = 1 A.O.S.  y = -1 Find the latus rectum The length of the l.r. is | 1/a | Since a = ¼ , the l.r. =4 Plot and label everything X = 1 V(2,-1) Y = -1 F(3,-1)

Example Identify the focus and directrix of the graph of the equation x = – y2. 1 8 The parabola is of the form x = ay2, so the vertex is at the origin and the parabola has a horizontal axis of symmetry. Since a < 0, the parabola opens to the left. | a | = 1 4c | – | = 1 4c 8 4c = 8 c = 2 The focus is at (–2, 0). The equation of the directrix is x = 2.

Example Identify the vertex, the focus, and the directrix of the graph of the equation x2 + 4x + 8y – 4 = 0. Then graph the parabola. x2 + 4x + 8y – 4 = 0 8y = –x2 – 4x + 4 Solve for y, since y is the only term. 8y = –(x2 + 4x + 4) + 4 + 4 Complete the square in x. y = – (x + 2)2 + 1 vertex form 1 8 The parabola is of the form y = a(x – h)2 + k, so the vertex is at (–2, 1) and the parabola has a vertical axis of symmetry. Since a < 0 (negative), the parabola opens downward.

Example (Cont) | a | = | – | = Substitute – for a. 4c = 8 Solve for c. 1 8 | a | = 4c = 8 Solve for c. c = 2 4c The vertex is at (–2, 1) and the focus is at (–2, –1). The equation of the directrix is y = 3. Locate one or more points on the parabola. Select a value for x such as –6. The point on the parabola with an x-value of –6 is (–6, –1). Use the symmetric nature of a parabola to find the corresponding point (2, –1).