How do you find the center, vertices, foci and eccentricity of $x^{2} + 4 y^{2} + 4 x + 32 y + 67 = 0$ $x^2+4y^2+4x+32y+67=0$ ?

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How do you find the center, vertices, foci and eccentricity of $x^{2} + 4 y^{2} + 4 x + 32 y + 67 = 0$ $x^2+4y^2+4x+32y+67=0$ ?

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Answer 1 · 2016-03-02T03:20:39

Ellipse. C: ( $-$ $-$ 2, $-$ $-$ 4 ); A, A', B and B' : ( $-$ $-$ 1, $-$ $-$ 4 ), ( $-$ $-$ 3, $-$ $-$ 4 ), ( $-$ $-$ 2, $-$ $-$ 3.5 ) and ( $-$ $-$ 2, $-$ $-$ 4.5 ); e = $\sqrt{}$ $sqrt$ 3 /2; S ( $-$ $-$ 2 + $\sqrt{}$ $sqrt$ 3 / 2, $-$ $-$ 4 ) and S' ( $-$ $-$ 2 $-$ $-$ $\sqrt{}$ $sqrt$ 3 / 2 ); ..

Explanation:

Rearranging to the standard form ( x $- α$ $- alpha$ )^2 / a^2 + ( y $- β$ $-beta$ )^2 / b^2 = 1, read semi-major axis a = 1, semi-minor axis b = 1/2, e = $\sqrt{}$ $sqrt$ ( 1 $-$ $-$ b^2/a^2 ) = $\sqrt{}$ $sqrt$ 3 / 2.
The rearranged form is
( x + 2 )^2 + ( y + 4 )^2/(1/4) = 1
This represents an ellipse with axes parallel to axes of coordinates.
Center is ( $α$ $alpha$ , $β$ $beta$ ) and foci are at ( $α$ $alpha$ $\pm$ $+-$ ae, $β$ $beta$ )

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How do you find the center, vertices, foci and eccentricity of $x^{2} + 4 y^{2} + 4 x + 32 y + 67 = 0$ $x^2+4y^2+4x+32y+67=0$ ?

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Explanation:

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How do you find the center, vertices, foci and eccentricity of x^2+4y^2+4x+32y+67=0x2+4y2+4x+32y+67=0x^2+4y^2+4x+32y+67=0?

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Explanation:

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How do you find the center, vertices, foci and eccentricity of $x^{2} + 4 y^{2} + 4 x + 32 y + 67 = 0$ $x^2+4y^2+4x+32y+67=0$ ?