Eccentricity of the ellipse 4{x^2} + {y^2} - | vedclass

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Question

(b) $4{x^2} - 8x + {y^2} + 2y + 1 = 0$

==> ${(2x - 2)^2} + {(y + 1)^2} = - 1 + 4 + 1$

==> $\frac{{{{(x - 1)}^2}}}{1} + \frac{{{{(y + 1)}^2

Vedclass · Accepted Answer

$\sqrt 3 /2$

Vedclass · Answer

(b) $4{x^2} - 8x + {y^2} + 2y + 1 = 0$

==> ${(2x - 2)^2} + {(y + 1)^2} = - 1 + 4 + 1$

==> $\frac{{{{(x - 1)}^2}}}{1} + \frac{{{{(y + 1)}^2}}}{4} = 1$

==> $e = \sqrt {1 - \frac{1}{4}} $

==> $e = \frac{{\sqrt 3 }}{2}$.

Eccentricity of the ellipse $4{x^2} + {y^2} - 8x + 2y + 1 = 0$ is

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