# 39 so that the equation of the hyperbola is y 1 5 2 x

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Chapter 11 / Exercise 11
Discrete Mathematics With Applications
Epp
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39 so that the equation of the hyperbola is y + 1 5 2 x 39 2 = 1 In Exercises 21–28, find the equation of the parabola with the given properties. 21. Vertex ( 0 , 0 ) , focus ( 1 12 , 0 ) solution Since the focus is on the x -axis rather than the y -axis, and the vertex is ( 0 , 0 ) , the equation is x = 1 4 c y 2 . The focus is ( 0 , c) with c = 1 12 , so the equation is x = 1 4 · 1 12 y 2 = 3 y 2 22. Vertex ( 0 , 0 ) , focus ( 0 , 2 ) solution The vertex is at ( 0 , 0 ) , so the equation is y = 1 4 c x 2 = 1 8 x 2 . 23. Vertex ( 0 , 0 ) , directrix y = − 5 solution The equation is y = 1 4 c x 2 . The directrix is y = − c with c = 5, hence y = 1 20 x 2 . 24. Vertex ( 3 , 4 ) , directrix y = − 2 solution If the graph were translated to the origin, the vertex would be ( 0 , 0 ) and the directrix would be translated down 4 units so would be y = − 6. Then c = 6 so the equation is y = 1 4 c x 2 = 1 24 x 2 . Translating back to ( 3 , 4 ) gives y = 1 24 (x 3 ) 2 + 4 25. Focus ( 0 , 4 ) , directrix y = − 4 solution The focus is ( 0 , c) with c = 4 and the directrix is y = − c with c = 4, hence the equation of the parabola is y = 1 4 c x 2 = x 2 16 . 26. Focus ( 0 , 4 ) , directrix y = 4 solution The focus is at ( 0 , c) with c = − 4 and the directrix is y = − c with c = − 4, hence the equation y = x 2 4 c of the parabola becomes y = − x 2 16 . Since c < 0, the parabola is open downward. 27. Focus ( 2 , 0 ) , directrix x = − 2 solution The focus is on the x -axis rather than on the y -axis and the directrix is a vertical line rather than horizontal as in the parabola in standard position. Therefore, the equation of the parabola is obtained by interchanging x and y in y = 1 4 c x 2 . Also, by the given information c = 2. Hence, x = 1 4 c y 2 = 1 4 · 2 y 2 or x = y 2 8 . 28. Focus ( 2 , 0 ) , vertex ( 2 , 0 ) solution The vertex is always midway between the focus and the directrix, so the directrix must be the vertical line x = 6, and c = − 2 2 = − 4. Since the directrix is a vertical line, the parabola is obtained by interchanging x and y in the equation for a parabola in standard position. Finally, c = − 2 2 = − 4 is the distance from the vertex to the focus, so the equation is x 2 = 1 4 c y 2 = − 1 16 y 2 , so x = 2 1 16 y 2
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Chapter 11 / Exercise 11
Discrete Mathematics With Applications
Epp
Expert Verified
May 26, 2011 S E C T I O N 12.5 Conic Sections 1493 In Exercises 29–38, find the vertices, foci, center (if an ellipse or a hyperbola), and asymptotes (if a hyperbola). 29. x 2 + 4 y 2 = 16 solution We first divide the equation by 16 to convert it to the equation in standard form: x 2 16 + 4 y 2 16 = 1 x 2 16 + y 2 4 = 1 x 4 2 + y 2 2 = 1 For this ellipse, a = 4 and b = 2 hence c = 4 2 2 2 = 12 3 . 5. Since a > b we have: The vertices are at ( ± 4 , 0 ) , ( 0 , ± 2 ) . The foci are F 1 = ( 3 . 5 , 0 ) and F 2 = ( 3 . 5 , 0 ) . The focal axis is the x -axis and the conjugate axis is the y -axis. The ellipse is centered at the origin. 30. 4 x 2 + y 2 = 16 solution We divide the equation by 16 to rewrite it in the standard form: 4 x 2 16 + y 2 16 = 1 x 2 4 + y 2 16 = 1 x 2 2 + y 4 2 = 1 This is the equation of an ellipse with a = 2, b = 4. Since a < b the focal axis is the y -axis. Also, c = 4 2 2 2 = 12 3 . 5. We get: The vertices are at ( ± 2 , 0 ) , ( 0 , ± 4 ) .