SOLUTION: Locate the center, foci, vertices, ends of latera recta, & draw the ellipse. also compute the eccentricity & find the equation of the directices. 1. 9x^2+25y^2=225

Question 948466: Locate the center, foci, vertices, ends of latera recta, & draw the ellipse. also compute the eccentricity & find the equation of the directices.
1. 9x^2+25y^2=225
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
Divide each side by 225

STANDARD FORM FOR ELLIPSE:
NOTE: The largest denominator is always a, the other b. Since the a is under the x term, the major axis is horizontal.
A). CENTER:Center is at (h,k) in this case (0,0)
B). FOCI:
(f) is distance from center to focus:
find the square root of each side
f=+4 or -4
ANSWER: The foci are at (4,0) and (-4,0)
C). VERTICES:END POINTS OF MAJOR AXIS:
+ or - a is the distance from center: a=5 so:
ANSWER: Vertices are at (5,0) and (-5,0)
D). END POINTS OF LATERA RECTA:
The latera recta are perpendicular to the major axis at the foci,
the length given by Since the focus is the midpoint, we use half this value , in this case (9/5)
ANSWERS:
The latus rectum at focus (4,0) has endpoints (4,9/5) and (4,-9/5)
The latus rectum at focus (-4,0) has endpoints (-4,9/5) and (-4,-9/5)
ECCENTRICITY:e=f/a=4/5 ANSWER e=eccentricity=4/5
EQUATIONS OF THE DIRECTRICES: 2a-f on major axis: 2a-f=10-4=6
ANSWERS:
Equation of directrix right of the origin: x=6
Equation of directrix left of the origin: x=-6

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