For Exercises 43–56, write the standard form of an equation of an ellipse subject to the given conditions. (See Example 5)
Vertices:
Foci:
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College Algebra (Collegiate Math)
- Re-arrange this vertex equation y= 2 (x + 1) - 6 in standard formarrow_forwardWhich are the foci for the hyperbola modeled by the equation (X-32 (x-1)²-1? 36 13 O (1. 10) and (1,-4) O (3.8) and (3, -8) O (1.9) and (1, -3) O (-6, 0) and (6, 0)arrow_forwardIn Exercises 51-60, convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 51. 9x? + 25y? – 36x + 50y – 164 = 0 52. 4x + 9y? - 32x + 36y + 64 = 0 53. 9x? + 16y? - 18x + 64y - 71 = 0 54. x + 4y? + 10x 8y + 13 = 0 55. 4x2 + y? + 16x – 6y – 39 = 0 56. 4x + 25y? – 24x + 100y + 36 = 0 57. 25x + 4y - 150x + 32y + 189 = 0 %3D 58. 49x? + 16y² + 98x – 64y - 671 = 0 59. 36x? + 9y2 - 216x = 0 %3! 60. 16x? + 25y² – 300y + 500 = 0 %3Darrow_forward
- 3) Explain how changing the a, h, and k variables affect the shape of the parabola in the vertex form y = a(x - h)² + k. [3 Communication Marks] Page 1arrow_forward2. Suppose that the last three digits of your student number are like. abc . It is not important whether it is a 4 digit number or a 5 digit number, take the last three :+(-1) ; (b+1)3 digits as abc. Consider the conic (-1)ª following. Major semiaxis Minor semiaxis (-1)ab. Find the %3D (a+1)a i. ii. iii. Vertices iv. Foci Eccentricity Directrices Asymptotes (if it is a hyperbola) and the graph (geogebra or nice handmade) V. vi. vii. Now, consider the Quadratic Surface (-1)ª. (a+1)a +(-1) + (-1)° (b+1) =(-1)abe. (c+1)2 Find the following. Name of the surface viii. ix. Traces of the surface Nice Graph (in GeoGebra or handmade) х. For example, your professor's ID number is 0000005121. So, he will study the conic (-1) + (-1) =(-1)*2 which gives (-) + (-1)', (1+1) = (-1)12 which gives (-) + = 1. Also he will (2+1)3 4 study the quadratic surface (-1)' (1+1)2 + (-1)² - (2+1)a + (-1)'; :(-1)121 (1+1) which gives. (-) + + (-÷) = 1.arrow_forward2. Obtain the graph of the parabola y = -x² + 4x using transformations of the graph of y = x². Explain each step.arrow_forward
- In Exercises 17-30, find the standard form of the equation of each parabola satisfying the given conditions. 17. Focus: (7,0); Directrix: x = -7 18. Focus: (9,0); Directrix: x = -9 19. Focus: (-5,0); Directrix: x = 5 20. Focus: (-10, 0); Directrix: x = 10 21. Focus: (0, 15); Directrix: y = -15 22. Focus: (0,20); Directrix: y = -20 23. Focus: (0, –25); Directrix: y = 25 24. Focus: (0, -15); Directrix: y = 15 25. Vertex: (2, -3); Focus: (2, -5) 26. Vertex: (5, -2); Focus: (7, -2) 27. Focus: (3, 2); Directrix: x = -1 28. Focus: (2, 4); Directrix: x = -4 29. Focus: (-3, 4); Directrix: y = 2 30. Focus: (7, –1); Directrix: y = -9arrow_forwardWrite an equation for the ellipse centered at the origin with vertices located at (-12, 0) and (12, 0) and foci at (-8, 0) and (8, 0).arrow_forwardSketch parabola y = x2 – 10x + 16 by plotting a few points to left and right of the vertex. (Find vertex using Vx = -b/2a)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage