Question

The centre of the ellipse $\frac{(x+y-3{)}^2}{9}+\frac{(x-y+1{)}^2}{16}=1$, is

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

Answer 1

Answer: (D)

We have,
$\frac{(x+y-3{)}^2}{9}+\frac{(x-y+1{)}^2}{16}=1$
To determine the centre of ellipse, put
$x+y-3=0$
$\Rightarrow x+y=3.......(i)$
and $x-y+1=0$
$\Rightarrow x-y=-1........(ii)$
On adding Eqs. (i) and (ii), we get
$x=1$ and $y=2$
Hence, the centre of the ellipse is $(1,2)$.