SOLUTION: Dear Sir/Ma'am Please Help me solve this problem. Find the equation if the Ellipse with center at (2,3), Vertices at (2,9) and (2,-3) and Eccentricity of 2/3/ Identify the pa

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Dear Sir/Ma'am Please Help me solve this problem. Find the equation if the Ellipse with center at (2,3), Vertices at (2,9) and (2,-3) and Eccentricity of 2/3/ Identify the pa      Log On


   



Question 1181020: Dear Sir/Ma'am
Please Help me solve this problem.
Find the equation if the Ellipse with center at (2,3), Vertices at (2,9) and (2,-3) and Eccentricity of 2/3/ Identify the parts of the Ellipse and sketch the graph.
Parts of the Ellipse
1. Center
2. Foci
3. Vertices V1, V2
4. Co Vertices B1, B2
5. Endpoints of Latus Rectum E1, E2, E3, E4
6. Directrices
7. Eccentricity
8. Length of LR
9. Length of Major Axis
10. Length of Minor Axis
Thank you and GOD Bless
Sincerely Yours,
Lorna

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
equation:
%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1
given:
center at (2,3)=>h=2 and k=3
Vertices at (2,9) and (2,-3)
as you can see major axis is parallel to y-axis, so it is b,the ellipse is vertical
and distance between vertices is 2b=12
=>b=6
so far your equation is:
%28x-2%29%5E2%2Fa%5E2%2B%28y-3%29%5E2%2F6%5E2=1
given ccentricity of 2%2F3
The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (b).
2%2F3=c%2Fb
2%2F3=c%2F6
c=%282%2F3%296
c=4
now find minor axis a
a=sqrt%286%5E2-4%5E2%29

%28x-2%29%5E2%2F20%2B%28y-3%29%5E2%2F36=1

semimajor axis length : 6
semiminor axis length: sqrt%2820%294.5
major axis length : 12
minor axis length: 9
foci:
(h, k%2Bc ), (h, k-c )
=>(2, 3%2B4 ), (2, 3-4 )
=>(2, 7 ), (2, -1 )

foci:
(h, k-c) , (h, k%2Bc)
(2, 3-4) , (2, 3%2B4)
(2, -1) , (2, 7)
covertices:
(h+%2B+b, k) ,(h+-+b, k)
(2++%2B+sqrt%2820%29, 3), (2+-+sqrt%2820%29, 3)
≈(6.5, 3) , (-2.5, 3)

eccentricity: 2%2F3
directrices: since you have vertical ellipse, directrices are +k=3 from each y coordinate of the vertices
y=9%2B+3=12
y=-3-3=-6
First directrix: y=-6
Second directrix: y=12
Endpoints of Latus Rectum
The chord of the ellipse through its one focus and perpendicular to the major axis (or parallel to the directrix) is called the latus rectum of the ellipse.
since foci is at: (2, -1) , (2, 7)
first latus rectum: y=-1
substitute in ellipse formula and you get
x+=+-4%2F3 or x+=+16%2F3
two Endpoints of Latus Rectum are
E1=-4%2F3,-1 and E1=16%2F3,-1
second latus rectum: y=7+
x same as above
E3=-4%2F3,7 and E4=16%2F3,7