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Q.
The centre of the ellipse $\frac{(x+y-2)^{2}}{9}+\frac{(x-y)^{2}}{16}=1$ is:
Jharkhand CECEJharkhand CECE 2002
Solution:
If the equation of ellipse be
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,
then centre of ellipse be $(0,0)$.
Given equation of ellipse be
$\frac{(x+y-2)^{2}}{9}+\frac{(x-y)^{2}}{16}=1$
For finding a centre, we put
$x+y-2=0$ and $x-y=0$
$\Rightarrow y=1, x=1$
$\therefore $ Required centre is $(1,1)$.