SOLUTION: find: a. center b. vertices c. foci d. extremities of the minor axis e. equation of the directrices f. eccentricity g. graph of the ellipse 81x^2 + 4y^2

SOLUTION: find: a. center b. vertices c. foci d. extremities of the minor axis e. equation of the directrices f. eccentricity g. graph of the ellipse 81x^2 + 4y^2 - 324x + 24y + 324 = 0

Question 976438: find: a. center b. vertices c. foci d. extremities of the minor axis e. equation of the directrices f. eccentricity g. graph of the ellipse
81x^2 + 4y^2 - 324x + 24y + 324 = 0
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
81x^2 + 4y^2 - 324x + 24y + 324 = 0
rewrite
81x^2- 324 x + 4y^2 + 24 y =-324
factor
81(x^2-4x) + 4(y^2 + 6y)=-324
complete the square, adding the product of the coefficient and the last term to the right side.
81 (x^2-4x+4) +4(y^2 +6y +9)=-324+360=36
Divide by 36 and rewrite
(81/36) (x-2)^2 + (4/36) (y+3)^2=1
{[9(x-2)/]4}^2+ {(y+3)/9}^2=1
{(x-2)/((4/9)}^2 +{(y+3)/9}^2=1
center is at (2,-3)
Larger denominator gives a^2 so that is 9, and a=3
Smaller denominator is b^2, so that is (2/3). This is an ellipse with the major axis along the y-axis.
So vertices are at (2,0) and (2,-6)
The semi-minor vertices are at (1 1/3,-3) and (2 2/3, -3)
foci: b^2+c^2=a^2, or c^2=9-(4/9)=(77/9). c^2=sqrt(77)/3
foci will be (2,-3+sqrt(77)/3 ) and (2, -3 - sqrt (77)/3 )
eccentricity is [sqrt(a^2-b^2)]/a = sqrt (77)/3 / 3 = sqrt (77)/9
y=(+/-) (a/e) for equation of directrices: They are perpendicular to major axis.
y=(+/-) a^2/c=9/[sqrt(77)/9]
y= (+/-) 81/sqrt (77)

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