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Question

Eccentricity of the ellipse x2+2y2-2x+3y+2=0 is


A

12

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B

12

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C

122

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D

13

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Solution

The correct option is A

12


Explanation for the correct option:

Step: 1 Simplify the given equation into a general equation of ellipse.

x2+2y2-2x+3y+2=0

(x2-2×1×x+1)+2[y2+2×y×34+(34)2]+1-98=0

(x-1)2+2(y+34)2=18

(x-1)218+(y+34)2116=1

The general equation of ellipse is (x-h)2a2+(y-k)2b2=1

If we compare this equation, then we will get:

a2=18,b2=116.

Step2. Finding eccentricity.

Therefore, eccentricity(e)

e=1-b2a2=1-(116)(18)=1-816=12

e=12

Hence, Option (A) is the correct answer.


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