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  4. The centre of the ellipse 4x2 + y2 - 8x + 4y - 8 = 0 is

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

KEAMKEAM 2017

Solution:

We have,
$4 x^{2}+y^{2}-8 x+4 y-8=0$
$\Rightarrow \left(4 x^{2}-8 x\right)+\left(y^{2}+4 y\right)-8=0$
$\Rightarrow 4\left(x^{2}-2 x\right)+\left(y^{2}+4 y\right)-8=0$
$\Rightarrow 4\left[(x-1)^{2}-1\right]+\left[(y+2)^{2}-4\right]-8=0$
$\Rightarrow 4(x-1)^{2}-4+(y+2)^{2}-4-8=0$
$\Rightarrow 4(x-1)^{2}+(y+2)^{2}=16$
$\Rightarrow \frac{(x-1)^{2}}{4}+\frac{(y+2)^{2}}{16}=1$
$\therefore $ Centre $=(1,-2)$