Chapter 10 Conic Sections 10.1 – The Parabola and the Circle 10.2 – The Ellipse 10.3 – The Hyperbola 10.4 – Nonlinear Systems of Equations and Their Applications

10.1 The Parabola and the Circle Graph parabolas of the form x = a(y – k)2 + h. Find the distance and midpoint between two points. Graph circles of the form (x – h)2 + (y – k)2 = r2. Find the equation of a circle with a given center and radius. x2 + y2 + dx + ey + f = 0. Chapter 1 Outline

Identify and Describe the Conic Sections Conic Section: A curve in a plane that is the result of intersecting the plane with a cone. Types of conic sections include parabolas, circles, ellipses, and hyperbolas.

Equations of Parabolas Opening Left or Right The graph of an equation in the form x = a(y – k)2 + h is a parabola with vertex at (h, k). The parabola opens to the right if a > 0 and to the left if a < 0. The equation of the axis of symmetry is y = k. Vertex (h, k) Vertex (h, k) y = k y = k x = a(y – k)2 + h, where a > 0 x = a(y – k)2 + h, where a < 0

Recall from Section 8.5 Example 6: f(x) = ax2 +bx + C = a(x – h)2 + k Recall from Section 8.5 Example 6: Given the function f(x) = x2 – 6x + 8 = (x – 3)2 – 1, determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. Find the domain and range. a = 1 > 0 up V (3, 1) Axis: x = 3 x y (x, y) 2 3 4 6 8 –1 (0, 8) (2, 0) (3, –1) (4, 0) (6, 8) y-intercept x-intercept vertex Domain: (−∞,∞ ), Range: [−1,∞ ) x-intercept

f(x) = ax2 +bx + C = a(x – h)2 + k Recall Sec 8.5 Example 7: f(x) = ax2 +bx + C = a(x – h)2 + k Given f(x) = – x2 +8x – 12 = – (x – 4)2 + 4, determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. Find the domain and range. Solution a = – 1 < 0, Down V (4, 4) & Axis: x = 4 x y (x, y) 2 4 6 8 –12 (0, – 12) (2, 0) (4, 4) (6, 0) (8, – 12) y-intercept x-intercept vertex x-intercept Domain: (−∞,∞ ) & Range: (−∞,4]

Example 3: Determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. x = 2y2 + 4y – 1 Solution a = 2 > 0, Parabola opens to the right Let us rewrite the equation: x = a(y – k)2 + h. x = 2y2 + 4y – 1 x + 1 = 2y2 + 4y x + 1 = 2(y2 + 2y) x + 1 + 2 = 2(y2 + 2y + 1) x + 3 = 2(y + 1)2 x = 2(y + 1)2 – 3 V (–3, –1) Axis: y = –1. X-intercept: (– 1,0) Y-intercepts: X=0: 0 = 2y2 + 4y – 1: (0, 0.22), (0, –2.22) x y –1 –2 5 1 –3

Example 4: Sketch the graph of x = – 2(y + 4)2 – 1. V (– 1, – 4). a = – 2 → Parabola opens to the left. x y –33 –8 –1 –4 – V (– 1, – 4). Axis: y = –4. x-int: (– 33,0) y-int: (0,?) x=0: -2(y + 4)2 – 1 =0

Learn the Distance and Midpoint Formulas The horizontal distance between the two points (x1, y1) and (x2, y2), indicated by the blue dashed line is |x2 – x1| and the vertical distance indicated by the red dashed line is |y2 – y1|. We can use the Pythagorean theorem along with the square root principle to derive: The Distance Formula The distance, d, between two points with coordinates (x1, y1) and (x2, y2), can be found using the formula:

Example 5: Find the distance between (–3, 7) and (–11, 9). If the distance is an irrational number, also give a decimal approximation rounded to three places. Solution Note: It does not matter which pair is (x1, y1) and which is (x2, y2).

Learn the Distance and Midpoint Formulas Given any two points (x1, y1) and (x2, y2), the point halfway between the given points can be found by the midpoint formula: Example 6: Find the midpoint of the line segment whose endpoints are (–4, 3) and (6, 7).

Graph Circles with Centers at the Origin A circle is the set of points in a plane that are the same distance, called the radius, from a fixed point, called the center. Circle with Its Center at the Origin and Radius r Circle with Its Center at (h, k) and Radius r

Graph Circles with Centers at the Origin Example 7: x2 + y2 = 16 is a circle C(0, 0) & r=4 Example 8: x2 + y2 = 10 is a circle C(0, 0) & r= 10 ≈3.16

Graph Circles with Centers at (h, k) Example 9: Determine the equation of the circle shown below. C(-3, 2) & r=3

Graph Circles with Centers at (h, k) Example 10: Find the center and radius of the circle and draw the graph. (x + 2)2 + (y  1)2 = 9 Solution C(2, 1). 9 = r2, r = 3 4 points: (  2, 4), (  2,  2) (  5, 1), ( 1, 1 ) r = 3 (-2, 1)

Example 11: Write the equation of the circle in standard form. center: (–5, 3); radius 4 Solution: (x – h)2 + (y – k)2 = r2 (x  (5))2 + (y – 3)2 = 42 (x + 5)2 + (y – 3)2 = 16 Example 12: C(0, 0); r = 5 Solution: x2 + y2 = r2 x2 + y2 = 52 x2 + y2 = 25

Example 13: Find the center and radius of the circle and draw the graph. x2 + 8x + y2  6y + 16 = 0 Solution C(4, 3) & r = 3 4 points: (  4,0), (  4,6) (  7, 3), ( 1, 3 )

Summary: Distance & Midpoint Formulas Circles Parabolas

Example 1: Given determine which way the graph opens, find the vertex and axis of symmetry, and draw the graph. Solution a = 2, h = 2, and k = 3. a > 0, the parabola opens up. V (2, 3) Axis: x = 2 y-int (0, 5)

Note that for h = 0 and k = 0, the circle is centered at the origin Note that for h = 0 and k = 0, the circle is centered at the origin. Otherwise, the circle is translated |h| units horizontally and |k| units vertically.

The center is at (–1, 3) and the radius is 2. Find the center and the radius and then graph the circle x2 + y2 + 2x – 6y + 6 = 0. x2 + 2x + y2 – 6y = –6 x2 + 2x + 1 + y2 – 6y + 9 = –6 + 1 + 9 (–1, 3) x y (x + 1)2 + (y – 3)2 = 4 (x – (–1))2 + (y – 3)2 = 22 The center is at (–1, 3) and the radius is 2.