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  4. The centre of the ellipse ((x+y-3)2/9)+((x-y+1)2/16)=1, is

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

TS EAMCET 2015

Solution:

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 $(1,2)$.