Question

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

• A. (0, 2)
• B. (2, - 1)
• C. (2, 1)
• D. (1, 2)
• E. (1, - 2)

Answer 1

Answer: (E)

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