Question

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

• A. (0, 0)
• B. (1, 1)
• C. (1, 0)
• D. (0, 1,)

Answer 1

Answer: (B)

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 $(1,1)$.