How to find the center, vertices, foci and eccentricity of x^2/36+y^2/64=1?

chaomeinguyzyp3

chaomeinguyzyp3

Answered question

2023-02-18

How to find the center, vertices, foci and eccentricity of x 2 36 + y 2 64 = 1 ?

Answer & Explanation

Gerald Dickerson

Gerald Dickerson

Beginner2023-02-19Added 10 answers

x 2 36 + y 2 64 = 1
Compared to the standard ellipse equation x 2 a 2 + y 2 b 2 = 1 ,
a 2 = 36 , b 2 = 64 b = 8 > 6 = a
Therefore, the eqn. represents an Ellipse with Major axis along Y- axis & Minor Axis along X-axis
The Eccentricity e is given by,
a 2 = b 2 ( 1 - e 2 ) 36 = 64 ( 1 - e 2 ) 36 64 = 1 - e 2
e 2 = 1 - 36 64 = 1 - 9 16 = 7 16 e = 7 4
The Focii are S ( 0 , b e ) = S ( 0 , 8 7 4 ) = S ( 0 , 2 7 ) &
S ( 0 , - b e ) = S ( 0 , - 2 7 )
The Centre of the ellipse is C ( 0 , 0 ) .
The Vertices are A ( a , 0 ) = A ( 6 , 0 ) , A ( - a , 0 ) = A ( - 6 , 0 ) , B ( 0 , b ) = B ( 0 , 8 ) , B ( 0 , - b ) = B ( 0 , - 8 )

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