SOLUTION: Dear Sir, Please help me solve this problem showing your solution. Find the equation if the ellipse with center at (2, 3), vertices at (2, 9) and (2, −3), and eccentricity

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Dear Sir, Please help me solve this problem showing your solution. Find the equation if the ellipse with center at (2, 3), vertices at (2, 9) and (2, −3), and eccentricity       Log On


   

Question 1181119: Dear Sir,
Please help me solve this problem showing your solution.
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.
Please help me also to identify the parts of the Ellipse and its sketch.
Parts of an Ellipse
1. Center
2. Foci
𝐹1
𝐹2
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 very much and God bless you.
Lorna
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the equation of the ellipse
%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1
given:
center at (2, 3),
since center is at (h, k)=>h=2, k=3
vertices at (2, 9) and (2, -3),
since vertices at (h, k%2Bb), (h, k-b)
3%2Bb=9=>b=6

=> since x coordinates same, major axis parallel y axis, and you have vertical ellipse
eccentricity of 2%2F3
e=sqrt%28b%5E2-a%5E2%29%2Fb+
2%2F3=sqrt%286%5E2-a%5E2%29%2F6
%282%2F3%29%5E2=%2836-a%5E2%29%2F6%5E2
%284%2F9%2936=%2836-a%5E2%29
%284%2F1%294=36-a%5E2
16=36-a%5E2
a%5E2=36-16
a%5E2=20
a=sqrt%2820%29
a=sqrt%284%2A5%29
a=2sqrt%285%29
c=sqrt%28b%5E2-a%5E2%29
c=sqrt%2836%2B20%29
c=sqrt%2816%29
c=4

so, your equation is
%28x-2%29%5E2%2F20%2B%28y-3%29%5E2%2F36=1




1. Center:(2, 3)
2. Foci: (h, k%2Bc ), (h,+k-c )
𝐹1 (2, 3%2B4 ) =>(2, 7 )
𝐹2 (2, 3-4 ) =>(2, -1 )
3. Vertices: (h, k%2Bb ), (h,+k-b )
𝑉1 (2, 3%2B6 ) =>(2, 9 )
𝑉2(2,+3-6 ) =>(2, -3 )
4. Co-vertices (h%2Ba, k ), (h-a,+k )
𝐵1(2%2B2sqrt%285%29, 3 )=>(6.5, 3 )
𝐵2(2-2sqrt%285%29, 3 )=>(-2.5, 3 )
5. Endpoints of Latus Rectum
foci is (2, 7 ), endpoints E1 and E2 lie on a line y=7
substitute in equation of ellipse to calculate x coordinates
%28x-2%29%5E2%2F20%2B%287-3%29%5E2%2F36=1...solve it and you will get
x+=+-4%2F3 and x+=+16%2F3
so,
E1 (-4%2F3, 7)
E2(16%2F3, 7)
other focus is at (2, -1 ),endpoints Ee and E4 lie on a line y=-1
x+=+-4%2F3 and x+=+16%2F3

E3 =>(-4%2F3, -1)
E4 =>(16%2F3,-1)
6. Directrices: y=-6, y=12}
7. Eccentricity:2%2F3
8. Length of LR: a%5E2%2Fk=20%2F3≈ 6.66666666666667
9. Length of Major Axis:12
10.Length of Minor Axis:2%2A2sqrt%285%29=4sqrt%285%29=8.9