Question

find the eccentricity and the foci of the ellipse (x^2/5) +(y^2/2) =1...​

Answer 1

Answer:

x^2 /5 +y^2 /2 =1

Step-by-step explanation:

a= √5

b= √2

c^2 = a^2 - b^2

= 5- 2

= 3

c = √3

foci = ( + √3 , 0)

eccentricity = ae = c

e = √3 /5

Answer 2

Answer:

focci(+- √6/√5,0)

eccentricity= √3/√5)

Step-by-step explanation:

see attachment

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