Review - Conic Sections PART I: Show All Work (on a separate sheet of paper): For 1 5: Write the standard form equation of each conic section using the given – information: A parabola whose focus is at (2, 4) and equation of the directrix is x = 6. A hyperbola whose center is at (2, 4), length of vertical transverse axis is 24, and length of conjugate axis is 18. An ellipse whose center is at (-2, -4), length of vertical major axis is 24, and length of minor axis is 18. A hyperbola whose center is at the origin, length of horizontal conjugate axis is 8, and distance of foci from center is 5. An ellipse whose center is at the origin, length of horizontal major axis is 16, and distance between foci is 12. For 6 12: Given the general form equation: – (a) identify the conic section represented (b) write the equation in standard form (c) find the necessary coordinates and equations relevant to that conic section (d) graph the equation (on graph paper) x2 4y2 4x 24y 36 = 0 – – – – x2 + 8x + 8y = 0 x2 + y2 + 12x + 10y + 45 = 0 16x2 + 4y2 + 96x 16y + 96 = 0 – 16x2 + 9y2 192x 72y + 576 = 0 – – 25y2 9x2 225 = 0 – – y2 8x 4y 20 = 0 – – –
Review - Conic Sections PART I: Show All Work (on a separate sheet of paper): For 1 – 5: Write the standard form equation of each conic section using the given information: A parabola whose focus is at (2, 4) and equation of the directrix is x = 6. A hyperbola whose center is at (2, 4), length of vertical transverse axis is 24, and length of conjugate axis is 18. An ellipse whose center is at (-2, -4), length of vertical major axis is 24, and length of minor axis is 18. A hyperbola whose center is at the origin, length of horizontal conjugate axis is 8, and distance of foci from center is 5. An ellipse whose center is at the origin, length of horizontal major axis is 16, and distance between foci is 12. For 6 – 12: Given the general form equation: (a) identify the conic section represented (b) write the equation in standard form (c) find the necessary coordinates and equations relevant to that conic section (d) graph the equation (on graph paper) x2 – 4y2 – 4x – 24y – 36 = 0 x2 + 8x + 8y = 0 x2 + y2 + 12x + 10y + 45 = 0 16x2 + 4y2 + 96x – 16y + 96 = 0 16x2 + 9y2 – 192x – 72y + 576 = 0 25y2 – 9x2 – 225 = 0 y2 – 8x – 4y – 20 = 0
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