Chapter 8 Analytical Geometry Copyright © 2014, 2010 Pearson Education, Inc.

Section 8.2 The Parabola Define a parabola geometrically. Find an equation of a parabola. Translate a parabola. Use the reflecting property of parabolas.

PARABOLA Let l be a line and F a point in the plane not on the line l. The the set of all points P in the plane that are the same distance from F as they are from the line l is called a parabola. Thus, a parabola is the set of all points P for which d(F, P) = d(P, l), where d(P, l) denotes the distance between P and l.

PARABOLA Point F is the focus. Line l is the directrix. Point F is the focus. the directrix is the axis or axis of symmetry. The line through the focus, perpendicular to The point at which the axis intersects the parabola is the vertex.

EQUATION OF A PARABOLA The equation y2 = 4ax is called the standard equation of a parabola with vertex (0, 0) and focus (a, 0). Similarly, if the focus of a parabola is placed on the negative x-axis, we obtain the equation y2 = – 4ax as the standard equation of a parabola with vertex (0, 0) and focus (–a, 0).

EQUATION OF A PARABOLA By interchanging the roles of x and y, we find that the equation x2 = 4ay is the standard equation of a parabola with vertex (0, 0) and focus (0, a). Similarly, if the focus of a parabola is placed on the negative x-axis, we obtain the equation x2 = – 4ay as the standard equation of a parabola with vertex (0, 0) and focus (0, –a).

MAIN FACTS ABOUT A PARABOLA WITH a > 0

MAIN FACTS ABOUT A PARABOLA WITH a > 0

MAIN FACTS ABOUT A PARABOLA WITH a > 0

MAIN FACTS ABOUT A PARABOLA WITH a > 0

MAIN FACTS ABOUT A PARABOLA WITH a > 0

MAIN FACTS ABOUT A PARABOLA WITH a > 0

Example: Graphing a Parabola Graph each parabola and specify the vertex, focus, directrix, and axis. a. x2 = −8y b. y2 = 5x a. The equation x2 = −8y has the standard form x2 = −4ay; so

Example: Graphing a Parabola The parabola opens down. The vertex is the origin, and the focus is (0, −2). The directrix is the horizontal line y = 2; the axis of the parabola is the y-axis. Since the focus is (0, −2) substitute y = −2 in the equation x2 = −8y of the parabola to obtain

Example: Graphing a Parabola Thus, the points (4, −2) and (−4, −2) are two symmetric points on the parabola to the right and the left of the focus.

Example: Graphing a Parabola b. The equation y2 = 5x has the standard form y2 = 4ax. The parabola opens to the right. The vertex is the origin, and the focus is The directrix is the vertical line the axis of the parabola is the x-axis.

Example: Graphing a Parabola To find two symmetric points on the parabola that are above and below the focus, substitute in the equation of the parabola.

Example: Graphing a Parabola Plot the two additional points and on the parabola.

LATUS RECTUM The line segment passing through the focus of a parabola, perpendicular to the axis, and having endpoints on the parabola is called the latus rectum of the parabola. The following figures show that the length of the latus rectum for the graphs of y2 = ±4ax and x2 = ±4ay for a > 0 is 4a.

LATUS RECTUM

LATUS RECTUM

Example: Finding the Equation of a Parabola Find the standard equation of a parabola with vertex (0, 0) and satisfying the given description. a. The focus is (–3, 0). b. The axis of the parabola is the y-axis, and the graph passes through the point (–4, 2). a. Vertex (0, 0) and focus (–3, 0) are both on the x-axis, so parabola opens left and the equation has the form y2 = – 4ax with a = 3. The equation is

Example: Finding the Equation of a Parabola b. Vertex is (0, 0), axis is the y-axis, and the point (–4, 2) is above the x-axis, so parabola opens up and the equation has the form x2 = – 4ay and x = –4 and y = 2 can be substituted in to obtain The equation is

Main facts about a parabola with vertex (h, k) and a > 0 Standard Equation (y – k)2 = 4a(x – h) Equation of axis y = k Description Opens right Vertex (h, k) Focus (h + a, k) Directrix x = h – a

Main facts about a parabola with vertex (h, k) and a > 0

Main facts about a parabola with vertex (h, k) and a > 0 Standard Equation (y – k)2 = –4a(x – h) Equation of axis y = k Description Opens left Vertex (h, k) Focus (h – a, k) Directrix x = h + a

Main facts about a parabola with vertex (h, k) and a > 0

Main facts about a parabola with vertex (h, k) and a > 0 Standard Equation (x – h)2 = 4a(y – k) Equation of axis x = h Description Opens up Vertex (h, k) Focus (h, k + a) Directrix y = k – a

Main facts about a parabola with vertex (h, k) and a > 0

Main facts about a parabola with vertex (h, k) and a > 0 Standard Equation (x – h)2 = – 4a(y – k) Equation of axis x = h Description Opens down Vertex (h, k) Focus (h, k – a) Directrix y = k + a

Main facts about a parabola with vertex (h, k) and a > 0

Example: Graphing a Parabola Find the vertex, focus, and the directrix of the parabola 2y2 – 8y – x + 7 = 0. Sketch the graph of the parabola. Complete the square on y.

Example: Graphing a Parabola Compare with the standard form We have h = –1, k = 2, and Vertex: (h, k) = (–1, 2) Focus: (h + a, k) = Directrix: x = h – a =

Example: Graphing a Parabola

REFLECTING PROPERTY OF PARABOLAS A property of parabolas that is useful in applications is the reflecting property: If a reflecting surface has parabolic cross sections with a common focus, then all light rays entering the surface parallel to the axis will be reflected through the focus. This property is used in reflecting telescopes and satellite antennas, because the light rays or radio waves bouncing off a parabolic surface are reflected to the focus, where they are collected and amplified. (See next slide.)

REFLECTING PROPERTY OF PARABOLAS

REFLECTING PROPERTY OF PARABOLAS Conversely, if a light source is located at the focus of a parabolic reflector, the reflected rays will form a beam parallel to the axis. This principle is used in flashlights, searchlights, and other such devices. (See next slide.)

REFLECTING PROPERTY OF PARABOLAS

Example: Calculating Some Properties of the Hubble Space Telescope The parabolic mirror used in the Hubble Space Telescope has a diameter of 94.5 inches. Find the equation of the parabola if its focus is 2304 inches from the vertex. What is the thickness of the mirror at the edges? (Not to scale)

Example: Calculating Some Properties of the Hubble Space Telescope Position the parabola so that its vertex is at the origin and its focus is on the positive y-axis. The equation of the parabola is of the form

Example: Calculating Some Properties of the Hubble Space Telescope To find the thickness y of the mirror at the edge, substitute x = 47.25 (half the diameter) in the equation x2 = 9216y and solve for y. Thus, the thickness of the mirror at the edges is approximately 0.242248 inch.

Section 8.3 The Ellipse Define an ellipse. Find an equation of an ellipse. Translate ellipses. Use ellipses in applications.

ELLIPSE An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a constant. The fixed points are called the foci (the plural of focus) of the ellipse.

ELLIPSE

EQUATION OF AN ELLIPSE is the standard form of the equation of an ellipse with center (0, 0) and foci (–c, 0) and (c, 0), where b2 = a2 – c2.

EQUATION OF AN ELLIPSE Similarly, by reversing the roles of x and y, an equation of the ellipse with center (0, 0) and foci (0, −c) and (0, c) and on the y-axis is given by .

HORIZONTAL AND VERTICAL ELLIPSES If the major axis of an ellipse is along or parallel to the x-axis, the ellipse is called a horizontal ellipse, while an ellipse with major axis along or parallel to the y-axis is called a vertical ellipse.

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)

Example: Finding an Equations of an Ellipse Find the standard form of the equation of the ellipse that has vertex (5, 0) and foci (±4, 0). Since the foci are (−4, 0) and (4, 0) the major axis is on the x-axis. We know c = 4 and a = 5; find b2. Substituting into the standard equation, we get .

Example: Graphing an Ellipse Sketch a graph of the ellipse whose equation is 9x2 + 4y2 = 36. Find the foci of the ellipse. First, write the equation in standard form:

Example: Graphing an Ellipse Because the denominator in the y2-term is larger than the denominator in the x2-term, the ellipse is a vertical ellipse. Here a2 = 9 and b2 = 4, so c2 = a2 – b2 = 5. Vertices: (0, ±3) Foci: Length of major axis: 6 Length of minor axis: 4

TRANSLATIONS OF ELLIPSES Horizontal and vertical shifts can be used to obtain the graph of an ellipse whose equation is The center of such an ellipse is (h, k), and its major axis is parallel to a coordinate axis.

Main facts about horizontal ellipses with center (h, k) Standard Equation Center (h, k) Major axis along the line y = k Length of major axis 2a Minor axis along the line x = h Length of minor axis 2b

Main facts about horizontal ellipses with center (h, k) Vertices (h + a, k), (h – a, k) Endpoints of minor axis (h, k – b), (h, k + b) Foci (h + c, k), (h – c, k) Equation involving a, b, and c c2 = a2 – b2 Symmetry The graph is symmetric about the lines x = h and y = k.

Graphs of horizontal ellipses

Main facts about vertical ellipses with center (h, k) Standard Equation Center (h, k) Major axis along the line x = h Length major axis 2a Minor axis along the line y = k Length minor axis 2b

Main facts about vertical ellipses with center (h, k) Vertices (h, k + a), (h, k – a) Endpoints of minor axis (h – b, k), (h + b, k) Foci (h, k + c), (h, k – c) Equation involving a, b, c c2 = a2 – b2 Symmetry The graph is symmetric about the lines x = h and y = k

Graphs of vertical ellipses

Example: Finding the Equation of an Ellipse Find an equation of the ellipse that has foci (–3, 2) and (5, 2), and has a major axis of length 10. Foci lie on the line y = 2, so horizontal ellipse. Center is midpoint of foci Length major axis =10, vertices at a distance of a = 5 units from the center. Foci at a distance of c = 4 units from the center.

Example: Finding the Equation of an Ellipse Use b2 = a2 – c2 to obtain b2. b2 = (5)2 – (4)2 = 25 – 16 = 9 to obtain b2. Major axis is horizontal so standard form is Replace: h = 1, k = 2, a2 = 25, b2 = 9 Center: (1, 2) a = 5, b = 3, c = 4

Example: Finding the Equation of an Ellipse Vertices: (h ± a, k) = (1 ± 5, 2) = (–4, 2) and (6, 2) Endpoints minor axis: (h, k ± b) = (1, 2 ± 3) = (1, –1) and (1, 5)

Example: Converting to Standard Form Find the center, vertices, and foci of the ellipse with equation 3x2 + 4y2 +12x – 8y – 32 = 0. Complete squares on x and y.

Example: Converting to Standard Form This is standard form. Center: (–2, 1), a2 = 16, b2 = 12, and c2 = a2 – b2 = 16 – 12 = 4. Thus, a = 4, and c = 2. Length of major axis is 2a = 8. Length of minor axis is

Example: Converting to Standard Form Center: (h, k) = (–2, 1) Foci: (h ± c, k) = (–2 ± 2, 1) = (–4, 1) and (0, 1) Vertices: (h ± a, k) = (–2 ± 4, 1) = (–6, 1) and (2, 1) Endpoints of minor axis:

Example: Converting to Standard Form

APPLICATIONS OF ELLIPSES 1. The orbits of the planets are ellipses with the sun at one focus. 2. Newton reasoned that comets move in elliptical orbits about the sun. 3. We can calculate the distance traveled by a planet in one orbit around the sun. 4. The reflecting property for an ellipse says that a ray of light originating at one focus will be reflected to the other focus.

REFLECTING PROPERTY OF ELLIPSES

Example: Lithotripsy An elliptical water tank has a major axis of length 6 feet and a minor axis of length 4 feet. The source of high-energy shock waves from a lithotripter is placed at one focus of the tank. To smash the kidney stone of a patient, how far should the stone be positioned from the source?

Example: Lithotripsy Since the length of the major axis of the ellipse is 6 feet, we have 2a = 6; so a = 3. Similarly, the minor axis of 4 feet gives 2b = 4 or b = 2. To find c, we use the equation c2 = a2 – b2. We have c2 = 32 – 22 = 5. Therefore,

Example: Lithotripsy If we position the center of ellipse at (0, 0) and the major axis along the x-axis, then the foci of the ellipse are and . The distance between these foci is ≈ 4.472 feet. The kidney stone should be positioned 4.472 feet from the source of the shock waves.

Section 8.4 The Hyperbola Define a hyperbola. Find the asymptotes of a hyperbola. Graph a hyperbola. Translate hyperbolas.

HYPERBOLA A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed points is constant. The fixed points are called the foci of the hyperbola.

HYPERBOLA Here is a hyperbola in standard position, with foci F1(–c, 0) and F2(c, 0) on the x-axis at equal distances from the origin. The two parts of the hyperbola are called branches.

EQUATION OF A HYPERBOLA is called the standard form of the equation of a hyperbola with center (0, 0). The x-intercepts are −a and a. The points corresponding to these x-intercepts are the vertices of the hyperbola.

PARTS OF A HYPERBOLA The line segment joining the two vertices is the transverse axis. The center is the midpoint of the transverse axis. The line segment joining (0, –b) and (0, b) is the conjugate axis.

EQUATION OF A HYPERBOLA Similarly, an equation of a hyperbola with center (0, 0) and foci (0, –c) and (0, c) on the y-axis is given by: Here the vertices are (0, − a) and (0, a). The transverse axis of length 2a of the graph of the equation lies on the y-axis, and its conjugate axis is the segment joining the points (−b, 0) and (0, b) of length 2b that lies on the x-axis.

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)

Example: Determining the Orientation of a Hyperbola Does the hyperbola have its transverse axis on the x-axis or y-axis? The orientation (left–right branches or up–down branches) of a hyperbola is determined by noting where the minus sign occurs in the standard equation. In the equation in this example, the minus sign precedes the y2-term, so the transverse axis is on the x-axis.

Example: Finding the Vertices and Foci from the Equation of a Hyperbola Find the vertices and foci for the hyperbola 28y2 – 36x2 = 63. First, convert the equation to the standard form.

Example: Finding the Vertices and Foci from the Equation of a Hyperbola Since the coefficient of x2 is negative, the transverse axis lies on the y-axis. and and therefore

Example: Finding the Vertices and Foci from the Equation of a Hyperbola Since we require a > 0 and c > 0, and c = 2. The vertices of the hyperbola are and The foci of the hyperbola are (0, −2) and (0, 2).

Example: Finding the Equation of a Hyperbola Find the standard form of the equation of a hyperbola with vertices (±4, 0) and foci (±5, 0). Since the foci of the hyperbola are on the x-axis, the transverse axis lies on the x-axis. The center of the hyperbola is midway between the foci, at (0, 0). The standard form of such a hyperbola is

Example: Finding the Equation of a Hyperbola The distance a between the center (0, 0) to either vertex, (−4, 0) or (4, 0) is 4; so a = 4 and a2 = 16. The distance c between the center to either focus, (5, 0) or (−5, 0) is 5; so c = 5 and c2 = 25. Use b2 = c2 – a2: b2 = 25 − 16 = 9. Substitute a2 = 16 and b2 = 9 into the standard form to get

THE ASYMPTOTES OF A HYPERBOLA WITH CENTER (0, 0) 1. The graph of the hyperbola has transverse axis along the x-axis and has the following two asymptotes:

THE ASYMPTOTES OF A HYPERBOLA WITH CENTER (0, 0) 2. The graph of the hyperbola has transverse axis along the y-axis and has the following two asymptotes:

Example: Finding the Asymptotes of a Hyperbola Determine the asymptotes of each hyperbola. a. The hyperbola is of the form so a = 2 and b = 3. Substituting these values into and we get the asymptotes

Example: Finding the Asymptotes of a Hyperbola b. The hyperbola has the form so a = 3 and b = 4. Substituting these values into and we get the asymptotes

PROCEDURE FOR GRAPHING A HYPERBOLA CENTERED AT (0, 0) Step 1 Write the equation in standard form. Determine transverse axis and the orientation of the hyperbola. Step 2 Locate vertices and the endpoints of the conjugate axis. Step 3 Lightly sketch the fundamental rectangle by drawing dashed lines parallel to the coordinate axes through the points in Step 2.

PROCEDURE FOR GRAPHING A HYPERBOLA CENTERED AT (0, 0) Step 4 Sketch the asymptotes. Extend the diagonals of the fundamental rectangle. These are the asymptotes. Step 5 Sketch the graph. Draw both branches of the hyperbola through the vertices, approaching the asymptotes. The foci are located on the transverse axis, c units from the center, where c2 = a2 + b2.

MAIN PROPERTIES OF HYPERBOLAS CENTERED AT (h, k) Standard Equation a > 0, b > 0 Equation of transverse axis y = k Length of transverse axis 2a Equation of conjugate axis x = h Length of conjugate axis 2b Center (h, k)

MAIN PROPERTIES OF HYPERBOLAS CENTERED AT (h, k) Vertices (h − a, k), (h + a, k) Endpoints of conjugate axis (h, k − b), (h, k + b) Foci (h − c, k), (h + c, k) Equation involving a, b, and c c2 = a2 + b2 Asymptotes

MAIN PROPERTIES OF HYPERBOLAS CENTERED AT (h, k) Standard Equation a > 0, b > 0 Equation of transverse axis x = h Length of transverse axis 2a Equation of conjugate axis y = k Length of conjugate axis 2b Center (h, k)

MAIN PROPERTIES OF HYPERBOLAS CENTERED AT (h, k) Vertices (h, k – a), (h, k + a) Endpoints of conjugate axis (h – b, k), (h + b, k) Foci (h, k – c), (h, k + c) Equation involving a, b, and c c2 = a2 + b2 Asymptotes

PROCEDURE FOR SKETCHING THE GRAPH OF A HYPERBOLA CETNERED AT (h, k) Step 1 Plot the center (h, k), and draw horizontal and vertical dashed lines through the center. Step 2 Locate the vertices and the endpoints of the conjugate axis. Lightly sketch the fundamental rectangle, with sides parallel to the coordinate axes, through these points.

PROCEDURE FOR SKETCHING THE GRAPH OF A HYPERBOLA CETNERED AT (h, k) Step 3 Sketch dashed lines through opposite vertices of the fundamental rectangle. These are the asymptotes. Step 4 Draw both branches of the hyperbola, through the vertices and approaching the asymptotes. Step 5 The foci are located on the transverse axis, c units from the center, where c2 = a2 + b2.

Example: Graphing a Hyperbola Show that 9x2 – 16y2 + 18x + 64y –199 = 0 is an equation of a hyperbola, and then graph the hyperbola. Complete the squares on x and y.

Example: Graphing a Hyperbola Steps 1-2 Locate the vertices. Center (–1, 2); a2 = 16, a = 4; b2 = 9, b = 3 (h – a, k) = (–1– 4, 2) = (–5, 2) (h + a, k) = (–1+ 4, 2) = (3, 2) Vertices: (3, –1), (3, 5), (–5, 5), (–5, –1). Draw the fundamental rectangle.

Example: Graphing a Hyperbola Step 3 Sketch the asymptotes. Extend the diagonals of the rectangle obtained in Step 2 to sketch the asymptotes: Step 4 Sketch the graph. Draw two branches opening left and right, starting from the vertices (–5, 2) and (3, 2) and approaching the asymptotes.

Example: Graphing a Hyperbola