# Chapter 10: Topics in Analytic Geometry

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1 Chapter 10: Topics in Analytic Geometry 10.1 Parabolas V In blue we see the parabola. It may be defined as the locus of points in the plane that a equidistant from a fixed point (F, the focus) and a fixed line ( d, the directrix). Note FP = TP. The axis or axis of symmetry of the parabola above is the y-axis. The focus is always on this axis and it is always perpendicular to the directrix. The segment through F and perpendicular to the axis is the latus rectum or focal width of the parabola. The point of the parabola midway between focus and directrix is called the vertex (V). If we take the vertex of the parabola to be at the origin (0, 0), the axis vertical as above, the focus at F(0, p), the directrix y = p, and focal width = 4p, we find the equation of the parabola: x = 4 p ythe parabola opens up (as shown) if p > 0, and down if p < 0. If we take the vertex of the parabola to be at the origin (0, 0), the axis horizontal, the focus at F(p, 0), the directrix x = p, and focal width = 4p, we find the equation of the parabola: y = 4 p xthe parabola opens right if p > 0, and left if p < 0. 1

2 Find vertex, focus, directrix, symmetry axis, and latus rectum of x + 6y = 0. The equation is of the form x = 4px if we write x = 6y. Since p = 3/ < 0, this parabola opens downward. The vertex is at V(0, 0), the focus F(0, 3/), the directrix is the horizontal line y = 3/, the symmetry axis is the y-axis (x = 0), and the latus rectum or focal with is 6. The parabola is graphed below Find an equation for the parabola that has its vertex at the origin and opens upward with focus 5 units from vertex. The focus is p = 5 from vertex and the equation is: x = 0y Find an equation for the parabola that has its vertex at the origin and directrix x =. The directrix is vertical so the parabola is horizontal, opening away from the directrix or to the left. The distance from V to directrix is p so p = and the equation is: y = 8x

3 Find vertex, focus, directrix, symmetry axis, and latus rectum of y 4x = 0. Writing y = 4x and comparing to y = 4px, we see the vertex is at V(0, 0) and p = 1. From the form of the equation and positive p we note the parabola opens to the right. The focus (always inside the parabola) is F(1, 0), the directrix is the vertical line x = 1, the symmetry axis is the x-axis (y = 0), and the latus rectum or focal width is 4p =

4 10. Ellipses The ellipse is the set of all points in the plane the sum of whose distances from two fixed points (the foci) is a constant (a). Let the center of the horizontal ellipse above be at (0, 0). The major axis (long axis) of the ellipse is then the horizontal segment stretching through the foci from one side of the ellipse to the other. The major axis intersects the ellipse at the vertices (V). The length of the major axis is a. The minor axis is the shorter axis of the ellipse, passing through the center and perpendicular to the major axis. It intersects the ellipse at the minor axis end-points B. The length of the minor axis is b. Each focus is a distance c (linear eccentricity) from the center. The foci are on the major axis. c = a b for the ellipse. The latus rectum or focal width of the ellipse is the segment through each focus perpendicular to the major axis. Its length is b / a. The eccentricity of the ellipse is a measure of its deviation from circularity. Always for the ellipse 0 < e < 1, with e = c/a. As e 0, the ellipse is more like a circle; as e 1, the ellipse is elongated. If the ellipse has center (0, 0), is longer in the horizontal direction, has vertices V(±a, 0), endpoints of minor axis B(0, ±b), foci F(±c, 0), and latus rectum endpoints (±c, ±b /a): x a y + = 1 is the equation of the horizontal ellipse b If the ellipse has center (0, 0), is longer in the vertical direction, has vertices V(0, ±a), endpoints of minor axis B(±b, 0), foci F(0, ±c), and latus rectum endpoints (±b /a, ±c): y a x + = 1 is the equation of the vertical ellipse. b 4

5 An alternative definition for the ellipse may be given in terms of directrices, pictured above. The ellipse is the set of points in the plane such that the distance of a point from a focus is e times the distance of the point from the associated directrix. The directrices above have equations x = ±a/e. Find V, B, F, e, length of major and minor axes, length of latera recta, endpoints of latera recta, and graph the ellipse below. x y + = 1 The form of this equation tells us the ellipse is horizontal, center origin. 5 9 a = 5, b = 9 (a is always the larger; if under x, the ellipse is horizontal) c = a b = 16 Thus a = 5, b = 3, c = 4. V(±5, 0) are endpoints of major axis. Major axis has length a = 10. B(0, ±3) are endpoints of minor axis. Minor axis has length b = 6. F(±4, 0) are coordinates of foci. They are always on major axis. e = c/a = 4/5 or 0.8 an elongated ellipse. Each latus rectum has length b /a = 18/5. This is the focal width of the ellipse. Endpoints of latera recta at (±4, ±9/5)

6 Find the equation of the ellipse with endpoints of major axis (±10, 0), distance between foci 6. The center is midway between V, B, F so here it is (0, 0). The major axis is horizontal, of length a = 10, so a = 5. The distance between foci is c = 6 so c = 3. Since c = a b, we find b = 5 3 = 16 and b = 4. Thus the equation: x y + = Find the equations of the ellipse with eccentricity 1/9, foci (0, ±). We see the center, midway between the foci, is at (0, 0). Since the foci are on the y-axis the ellipse is vertical. The distance from center to focus = c =. Now e = c/a 1/9 = /a a = 18. As above we use b = 18 = 30 and the desired equation is: y x + = Given 9x + 4y = 36, find: V, B, F, e, a, b, lengths and endpoints of latera recta, and graph The equation may be rewritten as: y x + = We see from its form that the center is (0, 0) and the ellipse is vertical. a = 9 and b = 4 c = 9 4 = 5 and a = 3, b =, and c = 5. Thus V(0, ±3), B(±, 0), F(0, ± 5), e = 5/3, length latus rectum = 8/5, and endpoints latera recta = (±4/5, ± 5)

7 10.3 Hyperbolas Hyperbolas may be defined as the set of points in the plane the difference of whose distances from two fixed points (the foci) is a constant (a). Alternatively, the distance of every point from a focus is e (the eccentricity) times greater than its distance from a fixed line associated with the focus called the directrix. We consider hyperbolas with center (0, 0). If the two branches of this curve open left and right (as pictured), then the distance between the vertices (transverse axis, VV') is a. Unlike the ellipse, this need not be the longer axis. The length of the conjugate axis BB' is b. The B points are not on the hyperbola, but help to define the rectangle shown above. The diagonals of the rectangle are the asymptotes, lines the hyperbola approach ever more closely as we get far from the origin. The foci are each c from the center, and located on the transverse axis. c = a + b The width of each branch of the hyperbola, measured through each focus, is the latus rectum or focal width. Each latus rectum has length b /a. The eccentricity e = c/a >1 fro the hyperbola. The asymptotes have slope ±b/a if the hyperbola opens left and right, ±a/b if it opens up and down. The equations of the asymptotes are y = (slope)x when hyperbola center at (0, 0). If the hyperbola has center (0, 0), opens left and right, has endpoints of transverse axis V(±a, 0), endpoints of conjugate axis B(0, ±b), foci F(±c, 0), latera recta endpoints (±c, ±b /a), and asymptotes y = ±(b/a)x, then it has equation: x y = 1 a b If the hyperbola has center (0, 0), opens up and down, has endpoints of transverse axis V(0, ±a), endpoints of conjugate axis B(±b, 0), foci F(0, ±c), latera recta endpoints (±b /a, ±c), and asymptotes y = ±(a/b)x, then it has equation: y x = 1 Note the + term is a. It may or may not be larger than b a b 7

8 Find V, B, F, e, length of transverse and conjugate axes, lengths of latera recta and their endpoints, equations of the asymptotes, and graph the hyperbola below. y x = 1 The equation implies center (0, 0), opens up and down 9 5 a = 9 (the + term - under y so a vertical hyperbola), b = 5, c = a + b = 34 Thus a = 3, b = 5, c = 34 Thus V(0, ±3), B(±5, 0), F(0, ± 34), e = ( 34)/3, transverse = 6, conjugate = 10, latera recta = 50/3, endpoints of latera recta (±5/3, ± 34), asy y = ±(3/5)x Write the equation for the hyperbola with foci (±5, 0), transverse axis = 6 The foci on the x-axis tell us the hyperbola opens left and right. The center is midway between the foci and is (0, 0). The foci are c from the center so c = 5. The length of the transverse axis is a = 6, so a = 3. Since c = a + b, we find b = 5 3 = 16. The desired equation is: x y =

9 10.4 Shifted Conics The center of parabolas, ellipses, and hyperbolas may not be at the origin. If the center is shifted horizontally and/or vertically to (h, k) then we replace x x h, and y y k. All features are measured from the center. Parabola: vertex (h, k), horizontal ( y k) = 4 p ( x h) Parabola vertex (h, k), vertical ( ) ( x h = 4 p y k) Ellipse center (h, k) horizontal Ellipse center (h, k) vertical Hyperbola center (h, k) horizontal Hyperbola center (h, k) vertical ( x h) ( y k) a + = 1 b ( y k) ( x h) a + = 1 b ( x h) ( y k) a = 1 b ( y k) ( x h) a = 1 b General Equation of a Shifted Conic The graph of the equation Ax + Cy + Dx + Ey + F = 0 where A and C are not both 0, is a conic or a degenerate conic. In the non-degenerate cases, the graph is: 1. Parabola is A or C is 0. Ellipse if A and C have same sign (or circle if A = C) 3. Hyperbola if A and C have opposite signs. Identifying Conics by the Discriminant (B 4 A C) The graph of the equation Ax + Bxy + Cy + Dx + Ey + F = 0, is a conic or a degenerate conic. In the non-degenerate cases, the graph is: 1. Parabola if B 4 A C = 0. Ellipse if B 4 A C < 0 3. Hyperbola if B 4 A C > 0 Find the equation of the hyperbola with one asymptote having equation y = x + 1, and vertices (ends of transverse axis) at (0, 0) and (0, ). The vertices are on a vertical transverse axis, so the hyperbola opens up and down. The center is midway between the vertices at (0, 1). The length of the transverse axis is = a so a = 1. The slope of the asymptote (a/b) = 1, so b = 1. ( ) ( ) ( ) y 1 x 0 y 1 x The equation is: = 1 =

10 Complete the square, determine the type of conic, find coordinates and lengths for all features, and graph 16x 9y 96x = 0. ( ) ( ) ( ) ( ) ( x 3) ( y 0) 16 x 6x + 9 y 0 = 0 16 x 6x y 0 = 144 = This is a hyperbola opening left and right, with center (3, 0) a = 9, b = 16, c = = 5 a = 3, b = 4, c = 5. V(0, 0)(6, 0) B(3, 4)(3, 4) F(, 0)(8, 0) e = 5/3 transverse axis length = 6, conjugate axis length = 8, latera recta length = 3/3 ends of latera recta(, ±16/3)(8, ±16/3), asy y 0 = ±(4/3)(x 3) Find vertex, focus, directrix, symmetry axis, focal width, and endpoints of latus rectum for the parabola: (x 3) = 8(y + 1) x vertical, vertex (3, 1), p> 0 so opens up, 4p = 8, so p =. Focus inside parabola so up from vertex or focus (3, 1). Symmetry axis is x = 3, directrix is y = 3. Focal width = 8 so ends of latus rectum at ( 1, 1) and (7, 1). 10

11 10.5 Rotation of Axes Given the most general form of second-degree equation below, reduce it to a simpler form by rotation of axes in order to find a conic horizontal or vertical relative to the new axes. Then use the results of previous sections to find components and graph. The most general form of second-degree equation is: Ax + Bxy + Cy +Dx + Ey + F = 0. We seek to find a positive, acute angle θ that will eliminate the xy term, giving: A'x' + C'y' +D'x' + E'y' + F' = 0, where the B' coefficient has been made zero. To do this, choose θ such that tan( θ) = B / (A C) if A C (else use θ = 45 ). Now replace every x and y in the original equation with: x = x' cos(θ) y' sin(θ) and y = x' sin(θ) + y' cos(θ) Given: 5x + 7xy +73y = 40x 30y + 75 Rotate by angle θ to eliminate x'y' term in new x'-y' coordinate system. Choose tan(θ) = B / (A C) = 7 / (5 73) = 7 / 1 = 4 / 7 Draw θ in quadrant I if positive, II if negative since θ is in interval (0, 180 ) We found the hypotenuse of 5 using the Pythagorean theorem. Observe cos(θ) = 7 / 5 and use the identities below. cos(θ) = 1+ cos(θ) = 3 / 5 sin(θ) = 1 cos(θ) = 4 / 5 Thus θ is a bit more than

12 Replace every x and y in the original expression with: x = x' cos(θ) y' sin(θ) and y = x' sin(θ) + y' cos(θ) x = x' y' and y = x' + y' Substituting, ( x' y' ) + 7( x' y' )( x' + y' ) + 73( x' + y' ) = ( x' y' ) 30( x' + y' ) Expanding, 1 [5(9x' 4x'y' + 16y' ) + 7(1x' 7x'y' 1y' ) + 73(16x' + 4x'y' + 9y' )] = 5 4x' 3y' 4x' 18y' + 75 Collecting like terms, x' ( ) + x'y'( ) + y' ( ) = 50y' Simplifying, 100x' + 5y' = 50y' + 75 or 4x' + y' = y' + 3 Translating (complete the square), 4x' + ( y' + y' + ) = 3 + or 4x' + (y' + 1) = 4 Writing in standard form, ( x' 0) ( y' + 1) = 1 An ellipse with center (0, 1) 1

13 Graphing, Maple plot: implicitplot(5*x^ + 7*x*y + 73*y^ = 40*x - 30*y + 75,x=-5..5,y=-5..5); Note: If one simply wishes to rotate axes by a given angle and transform single points to the new system, the inverse of the above transformations may be used. These are: x' = x cos(θ) + y sin(θ) and y' = x sin(θ) + y cos(θ) 13

14 10.6 Polar Equations of Conics ed ed r = or r = 1± e cos θ 1± e sin θ ( ) ( ) is a conic with one focus at origin and eccentricity e. The conic is a parabola is e < 1, an ellipse if 0 < e < 1, and a hyperbola if e > 1. x = 1 1 Pictured above is a graph of r =. This has e = 1 and d = cos( θ ) The focus of the parabola is at the pole, its vertex at (1/, 0 ). The directrix is to the right of the pole at x = 1 when using + in denominator. 1 If the equation were r =, the parabola would open to the right. 1 cos( θ ) The directrix would be x = 1. The focus would still be at the pole, the vertex now at (1/, 180 ). 1 If the equation were r =, the parabola would open down. 1 + sin( θ ) The directrix would be y = 1. (Note + directrix when + in denominator) The focus would still be at the pole, the vertex now at (1/, 90 ). 1 If the equation were r =, the parabola would open up. 1 sin( θ ) The directrix would be y = 1. The focus would still be at the pole, the vertex now at (1/, 70 ). 14

15 Example: Identify the conic 6 r = + sin θ We must put this in the form ( ) ed r = so divide num and denom by 1 + e sin θ 3 r = e = 1/ and d = 6. Since e = 1/, this is an ellipse sin( θ) Since there is a +sin(θ) in denominator, the ellipse is vertical, with its upper focus at the pole. To graph quickly, find r when θ = 0, 90, 180, and 70. ( )

16 10.7 Parametric Equations It is often useful to express x and y coordinates in terms of a parameter such as time t. We write x = f(t) and y = g(t) as the parametric equations of a curve. Sketch the curve defined by x = t, y = t, for t 4. One prepares a table of values of t and the corresponding x and y. Only the (x, y) values are plotted t x y Given parametric equations x = t, y = t, for t 4, eliminate the parameter to obtain a single equation in x and y. Solve for t in one of the equations and then substitute into the other. t = y x = (y ) or (y ) = x Recall (y k) = 4 p (x h) is the standard form of a shifted parabola that opens to the right with vertex at (h, k). Thus the graph should have vertex at (0, ) and open right. However, we must note t lies between and 4 meaning the curve and graph are only defined for 4 x 16 and 0 y. We draw this only. Note: To graph parametric equations with Mathematica, use ParametricPlot[ ]. On the TI-86, from the Home screen set Mode Param. Given x = sin(t) and y = 3 cos(t), eliminate the parameter. Square both equations: x = 4 sin (t) and y = 9 cos (t) or x y = sin () t and = cos () t Now add these equations. 4 9 x y + = sin () t + cos () t = 1 Vertical ellipse, center (0, 0)

17 Given x = sin(3 t) and y = sin(4 t), graph this Lissajous figure Note: Projectile motion equations are conveniently written in parametric form with time t as the parameter. If the initial velocity of a projectile is v 0 and it is launched at an angle θ above the horizontal, then we can resolve v 0 into vertical (v 0 sin(θ)) and horizontal (v 0 cos(θ)) components. In the absence of air resistance, and with gravity providing an acceleration (g = 3 ft/s = 9.80 m/s near the surface of the earth, directed downward) only in the y direction, we write x = x 0 + (v 0 cos(θ)) t and y = y 0 + (v 0 sin(θ)) t (1/) g t (x 0, y 0 ) is the starting position of the projectile. It is often taken to be (0, 0). Suppose a projectile is fired from position (0, 0) at t = 0 with an initial speed of 048 ft/s at an angle of 30 above the horizontal. What is the maximum height attained by the projectile? When and where will the projectile hit the ground? Graph. Eliminate the parameter t and show the path is a parabola. First we solve for the time the projectile is in the air. Assuming level ground, we find at what times the projectile is at y = 0: Solving 0 = 0 + (048 sin(30 )) t (1/) 3 t we find t = 0 and 64 s. The time up equals the time down for level ground so t = 3 s to max height. Maximum height = 0 + (048 sin(30 )) 3 (1/) 3 3 = 16,384 ft. Hits ground at t = 64 s and x = 0 + (048 cos(30 )) 64 = 113,51 ft 17

18 Now we must eliminate the parameter t and show the path is a parabola. Assume (x 0, y 0 ) is (0, 0). Solve for t in the x equation below and substitute. Thus x = (v 0 cos(θ)) t and y = (v 0 sin(θ)) t (1/) g t becomes: x x 1 x t = y= v0 sin( θ) g v0 cos( θ) v0 cos( θ) v0 cos( θ) We stop here as we have an equation of the form y = c 1 x c x, a parabola Note: Note: It is shown in the text that the range downfield for level ground may be written: v0 sin( θ) x = This agrees with the value for maximum x found above. g Polar equations may be put into parametric form. Recall: x = r cos(θ) and y = r sin(θ). Then if one is give r = f(θ), write: x = f(θ) cos(θ) and y = f(θ) sin(θ) Write r = sin(4θ) in parametric form. x = sin 4θ cos θ y = sin 4θ sin θ ( ) ( ) ( ) ( ) 18

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