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10.1 Circles and Parabolas

10.1 Circles and Parabolas. Conic Sections Parabolas, circles, ellipses, hyperbolas. 6.1 Circles. A circle is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center.

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10.1 Circles and Parabolas

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  1. 10.1 Circles and Parabolas • Conic Sections • Parabolas, circles, ellipses, hyperbolas

  2. 6.1 Circles A circle is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center. • A circle with center (h, k) and radius r has length to some point (x, y) on the circle. • Squaring both sides yields the center-radius form of the equation of a circle.

  3. 6.1 Finding the Equation of a Circle Example Find the center-radius form of the equation of a circle with radius 6 and center (–3, 4). Graph the circle and give the domain and range of the relation. Solution Substitute h = –3, k = 4, and r = 6 into the equation of a circle.

  4. 6.1 Graphing Circles with the Graphing Calculator Example Use the graphing calculator to graph the circle in a square viewing window. Solution

  5. 6.1 Graphing Circles with the Graphing Calculator • TECHNOLOGY NOTES: • Graphs in a nondecimal window may not be connected • Graphs in a rectangular (non-square) window look like an ellipse

  6. 6.1 Finding the Center and Radius of a Circle Example Find the center and radius of the circle with equation Solution Our goal is to obtain an equivalent equation of the form We complete the square in both x and y. The circle has center (3, –2) with radius 3.

  7. 6.1 Equations and Graphs of Parabolas A parabola is a set of points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line, the directrix. • For example, let the directrix be the line y = –p and the focus be the point F with coordinates (0, p).

  8. 6.1 Equations and Graphs of Parabolas • To get the equation of the set of points that are the same distance from the line y = –p and the point (0, p), choose a point P(x, y) on the parabola. The distance from the focus, F, to P, and the point on the directrix, D, to P, must have the same length.

  9. 6.1 Parabola with a Vertical Axis The parabola with focus (0, c) and directrix y = –p has equation x2 = 4py. The parabola has vertical axis x = 0, opens upward if p > 0, and opens downward if p < 0. • The focal chord through the focus and perpendicular to the axis of symmetry of a parabola has length |4p|. • Let y = p and solve for x. The endpoints of the chord are ( x, p), so the length is |4p|.

  10. 6.1 Parabola with a Horizontal Axis The parabola with focus (c,0) and directrix x = –p has equation y2 = 4px. The parabola has horizontal axis y = 0, opens to the right if p > 0, and to the left if p < 0. • Note: a parabola with a horizontal axis is not a function. • The graph can be obtained using a graphing calculator by solving y2 = 4px for y: Let and graph each half of the parabola.

  11. 6.1 Determining Information about Parabolas from Equations Example Find the focus, directrix, vertex, and axis of each parabola. (a) Solution (a) Since the x-term is squared, the parabola is vertical, with focus at (0, p) = (0, 2) and directrix y = –2. The vertex is (0, 0), and the axis is the y-axis.

  12. 6.1 Determining Information about Parabolas from Equations (b) The parabola is horizontal, with focus (–7, 0), directrix x = 7, vertex (0, 0), and x-axis as axis of the parabola. Since p is negative, the graph opens to the left.

  13. 6.1 Writing Equations of Parabolas Example Write an equation for the parabola with vertex (1, 3) and focus (–1, 3). Solution Focus lies left of the vertex implies the parabola has • a horizontal axis, and • opens to the left. Distance between vertex and focus is 1–(–1) = 2, so p = –2.

  14. 6.1 An Application of Parabolas Example Signals coming in parallel to the axis of a parabolic reflector are reflected to the focus, thus concentrating the signal. The Parkes radio telescope has a parabolic dish shape with diameter 210 feet and depth 32 feet.

  15. 6.1 An Application of Parabolas • Determine the equation describing the cross section. • The receiver must be placed at the focus of the parabola. How far from the vertex of the parabolic dish should the receiver be placed? Solution (a) The parabola will have the formy = ax2 (vertex at the origin) and pass through the point

  16. 6.1 An Application of Parabolas • Since The receiver should be placed at (0, 86.1), or 86.1 feet above the vertex.

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