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Mathematics 92 Online
OpenStudy (anonymous):

I will give medal : Find the equation of the ellipse given latus rectum 60/19, distance between directrices 2 sqrt 19

OpenStudy (anonymous):

@sillynikki1

OpenStudy (anonymous):

The length of each Latus Rectum is 2b²/a. The distance from the centre to a directrix is, d = a/e so the distance between the two directrices is 2a/e As e (the eccentricity) = √[(a² - b²)/a²] then 2a/e = 2a/√[(a² - b²)/a²] = 2a²/√(a² - b²). The general relationship between a,b and c is, b² = a² - c² so, c² = a² - b² and therefore c = √(a² - b²).

OpenStudy (anonymous):

a) 60/19 = 2b²/a : b² = 60a/38 = 30a/19 2√19 = 2a²/√(a² - b²) : √19 = a²/√(a² - b²) : √[19(a² - b²)] = a² : 19a² - 19b² = a^4 : b² = (19a² - a^4) / 19 Equating expressions for b² gives, 30a/19 = (19a² - a^4) / 19 : 30 = 19a - a³ : a³ - 19a + 30 = 0 then a = -5, 3 and 2. . . taking the positive solutions gives a² = 9 : a² = 4 b² = 30a/19 = 30*3/19 = 90/19 : b² = 30*2/19 = 60/19. The equations for these ellipses are : (1) [x²/9] + [y²/(90/19)] = 1 . . . 10x² + 19y² = 90 (2) [x²/4] + [y²/(60/19)] = 1 . . . 15x² + 19y² = 60

OpenStudy (anonymous):

b) 2c = (4/3)√33 = 2√(a² - b²) : √(a² - b²) = ⅔√33 : a² - b² = 132/9 = 44/3 . . . a² = b² + (44/3) Substituting the given coordinates into the general equation : [4/a²] + [1/b²] = 1 : 4b² + a² = a²b². Substituting for a² : 4b² + b² + (44/3) = [b² + (44/3)]b² = b^4 + (44/3)b². b^4 + (29/3)b² - (44/3) = 0 . . . solving for b² . . . . b² = (4/3) and -11. Taking the positive solution, b² = (4/3) thus a² = (4/3) + (44/3) = 16. Which provides the equation : [x²/16] + [y²/(4/3)] = 1 ... x² + 12y² = 16

OpenStudy (anonymous):

a³ - 19a + 30 = 0,how did you get a by this equation?

OpenStudy (anonymous):

@sillynikki1

OpenStudy (anonymous):

I actually can't remember how I got that... For some reason, once I answer a question, I can't remember how I got to the answer

OpenStudy (anonymous):

i cant solve it lol, i dont know how to work with it :D

OpenStudy (anonymous):

lol, I don't remember, but I'll look at it again

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