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

(0, 2)

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(2, - 1)

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(2, 1)

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(1, 2)

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Solution

The correct option is D $(1, 2)$
We have,

$4 x^{2} + y^{2} - 8 x + 4 y - 8 = 0$
$\Rightarrow (4 x^{2} - 8 x) + (y^{2} + 4 y) - 8 = 0$
$\Rightarrow (4 x^{2} - 2 x) + (y^{2} + 4 y) - 8 = 0$

$\Rightarrow 4 [(x - 1)^{2} - 1] + [(y + 2)^{2} - 4] - 8 = 0$
$\Rightarrow 4 (x - 1)^{2} - 4 + (y + 2)^{2} - 4 - 8 = 0$

$\Rightarrow 4 (x - 1)^{2} + (y + 2)^{2} = 16$

$\Rightarrow \frac{(x - 1)^{2}}{4} + \frac{(y + 2)^{2}}{16} = 1$

$∴$ Centre $= (1, 2)$

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Similar questions

4 x^{2} + y^{2} - 8 x + 2 y + 1 = 0

(1, 2)

x^{2} + y^{2} - 8 x + 4 y + 16 = 0

(- 1, 2)

4 x^{2} + y^{2} + 8 x - 4 y + 4 = 0

Then eccentricity of the Ellipse $4 x^{2} + y^{2} - 8 x - 2 y + 1$ = 0

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