FIND THE EQUATION OF ELLIPSE FROM VERTICES AND ENDPOINTS

Problem 1 :

Vertices: (5, 0), (5, 12)

endpoints of the minor axis: (1, 6), (9, 6)

Solution:

The symmetric is about y-axis. 

(x-h)2b2+(y-k)2a2=1

Minor axis = 2b

minor axis: (1, 6), (9, 6)

2b=(9-1)2+(6-6)22b=822b=8b=4

Major axis = 2a

Vertices: (5, 0), (5, 12)

2a=(5-5)2+(12-0)2=1222a=12a=6

The midpoint of vertices is the center of the ellipse

(h,k)=5+52,0+122=102,122(h,k)=(5,6)

The equation of the ellipse is

(x-5)242+(y-6)262=1(x-5)216+(y-6)236=1

Problem 2 :

Vertices: (0, 2), (4, 2)

endpoints of the minor axis: (2, 3), (2, 1)

Solution:

The symmetric is about x-axis. 

(x-h)2a2+(y-k)2b2=1

Minor axis = 2b

minor axis: (2, 3), (2, 1)

2b=(2-2)2+(1-3)22b=(-2)22b=42b=2b=1

Major axis = 2a

Vertices: (0, 2), (4, 2)

2a=(4-0)2+(2-2)2=422a=4a=2

The midpoint of vertices is the center of the ellipse

(h,k)=0+42,2+22=42,42(h,k)=(2,2)

The equation of the ellipse is

(x-2)222+(y-2)222=1(x-2)24+(y-6)21=1

Problem 3 :

Major axis vertical with length 10

Length of minor axis 4; Center (-2, 3)

Solution:

Major axis is vertical

(x-h)2b2+(y-k)2a2=1

Center (h, k) = (-2, 3)

Major axis = 2a

Minor axis = 2b

2a = 10

a = 5

2b = 4

b = 4

So, the equation of an ellipse is

(x+2)222+(y-3)252=1(x+2)24+(y-3)225=1

Problem 4 :

Endpoints of Major axis: (7, 9) & (7, 3)

Endpoints of Minor axis: (5, 6) & (9, 6)

Solution:

(x-h)2b2+(y-k)2a2=1

Major axis: (7, 9) (7, 3)

Center (h,k)=7+72,9+32=142,122(h,k)=(7,6)

Major axis = 2a

Minor axis = 2b

2a=(7-7)2+(3-9)2=(-6)22a=362a=6a=3
2b=(9-5)2+(6-6)22b=422b=162b=4b=2

So, equation of ellipse is

(x-7)222+(y-6)232=1(x-7)24+(y-6)29=1

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