x y 2 2 CONIC SECTIONS Problem 1

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1 CONIC SECTIONS Problem For the equations below, identify each conic section If it s a parabola, specify its vertex, focus and directrix If it s an ellipse, specify its center, vertices and foci If it s a hyperbola, specify its center, vertices, foci and asymptotes a) x y y 3 9 x y y 3 9 x y y x y x y x y 3 So the conic is an ELLIPSE, where a 3, b, and c 3 We therefore conclude the following: Its center is the point C 0, Its major axis is the horizontal line y and its minor axis is the y -axis Its vertices are located at the points 3, Its foci are located at the points, V and 3, F and F, V b) 8y 4 x 8y x x 4 y

2 So the conic is a PARABOLA that OPENS UP, where a We therefore conclude the following: Its vertex is located at the point V, 3 Its focus is located at the point F, 5 Its directrix is the horizontal line y 9 x 36 4y c) x y x y 3 So the conic is a HYPERBOLA, where conclude the following: a, b 3, and c We therefore Its center is the point C,0 Its transverse axis is the x-axis and its conjugate axis is the vertical line x Its vertices are located at the points V 3,0 and V,0 Its foci are located at the points F and 3, 0 F 3, Its asymptotes are the lines y x and y x d) x 3y 4x 6y 3 x x y y 3 3 x y x y

3 So the conic is an ELLIPSE, where a 3, the following: b 6, and c We therefore conclude Its center is the point C, Its major axis is the horizontal line y and its minor axis is the vertical line Its vertices are located at the points V 4, and V, Its foci are located at the points F 3, and F 3, x e) x y 3 x 8y x x y y 8 3 x y y x y y x 4 y x So the conic is a HYPERBOLA, where conclude the following: a, b, and c 4 6 We therefore Its center is the point C, Its transverse axis is the vertical line x and its conjugate axis is the horizontal line y Its vertices are located at the points V, and V, Its foci are located at the points F, 6 and F, 6 4 Its asymptotes are the lines y x 4 and y x

4 Problem [Exercise # 74 on page 645: Parabolic Arch Bridge] Suppose the parabolic arch is set on the x-y plane so that its vertex V is located on the positive y-axis at V 0,k and its two legs are located at the x-intercepts 50,0 and 50,0 Then, by assumption, the point 40,0 is also on the parabola We seek to find k Since this parabola opens DOWN and has vertex V 0,k, it has an equation of the form x 4ay k, where a 0 Plugging in the coordinates of the points 50,0 and 40, a0 k 4ak 40a yields the x system of equations, which implies that 500 4ak a 900 and thus results in the solution ak,, The height of the bridge at 9 50 its center is then given by 778 ft 9 Problem 3 [Exercise # 7 on page 655: Whispering Gallery] Here the whispering gallery is shaped like the upper portion of an ellipse whose foci are located 00 feet from each other and whose vertices are located 6 feet from the foci This implies that if we place the center of the ellipse at the origin, then the foci are located at the points 50,0 and the vertices are located at the points 56,0 We then have c 50, a 56, and b a c Therefore, the gallery is feet long and has a maximal height at the center of the room of approximately 5 feet Problem 4 [Exercise # 77 on page 655: Installing a Vent Pipe] Here the length of the major axis of this elliptical hole is the hypotenuse of a right triangle with base 8 inches and height 0 inches (since the diameter of the vent is 8 inches and the pitch 5 of the roof is ) It is then equal to inches, or approximately 8 4 inches The length of the minor axis of this elliptical hole is simply the diameter of the vent, or 8 inches

5 Problem 5 [Exercise # 78 on page 655: Volume of a Football] First, note that a longitudinal cross-section of the prolate spheroid (ie the football) passing through its center is an ellipse in the Cartesian plane whose center is at the origin, whose major axis is the x-axis, and whose vertices and foci lie on the x-axis Since the football is 5 inches in length, the vertices of the ellipse are located at the points 5565,0 and 5565,0, and so we have a 5565 Since the football has a center circumference of 85 inches, we have b 85 [do you see 85 why?] This implies that b 4496 We conclude that the volume of the football, in cubic inches, is given by V Problem 6 [Exercise # 8 p 669: Equilateral Hyperbola] We know that the eccentricity of an ellipse, denoted by e, is defined as the positive number c a We also know that if the hyperbola is equilateral, then a b Since b c a for any hyperbola, we conclude that equilateral hyperbola Therefore its eccentricity is given by c b a a a for an c a e a a

6 Problem 7 a) Identify the conic section represented by the polar equation r3 rcos r 3 rcos r rcos 3 3 r cos 3 3 r cos cos Therefore, e and the conic is a hyperbola b) See attached graph c) Find the rectangular equation of the conic Justify your work analytically If r 3rcos, then we can convert the polar equation of the hyperbola to rectangular coordinates as follows: r x y 3 x 3 rcos x y 9 x 4x y x x y x x y x y x x y 3 x 3 y

7 PARAMETRIC EQUATIONS Problem Consider the plane curve C defined by the following parametric equations: x 4sin t sin( t) y 4cost cos( t), (t R) a) See attached graph b) What type of graph do you recognize here? A cardioid c) Find the polar equation of C Justify your work analytically If x 4sin(t) sin(t) and y 4cos(t) cos(t), then we have r x y 6sin (t) cos (t) 4sin (t) cos (t) 6sin(t)sin(t) cos(t)cos(t) 6() 4() 6sin(t)sin(t) cos(t)cos(t) 0 6sin (t)cos(t) cos(t) sin (t) 0 6cos(t) Letting t be any real number, the polar equation of C is then given by r 0 6cos

8 Problem [Exercise # 54 p 696: Projectile Motion] 0 Here we have v0 5 ft / sec, 40, and h 3 ft a) The parametric equations that model the position of the ball are given by 0 0 x x( t) v cos t 5cos 40 t y y t gt v0 t h t t 0 ( ) sin 6 5sin 40 3 b) To find out how long the ball is in the air, find the positive root to the quadratic equation y 0: y 0 0 6t 5sin 40 t sin 40 5sin t 6 t 506sec So the ball traveled for a total of 506 seconds c) To determine the horizontal distance that the ball traveled, compute x (506) : 0 x(506) 5cos ft So the ball traveled an approximate total horizontal distance of 4845 feet d) To determine the maximal height attained by the ball, determine the y-coordinate of the b vertex point of the parabola y y() t, or y a : 0 b 5sin 40 y y y ft a ( 6) So the ball reached a maximal height of approximately 0387 feet