Question

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

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

Answer 1

Answer: (B)

The equation of the ellipse is
$9(x+y-2{)}^2+16(x-y{)}^2=144$ $\dots \dots (1)$
Differentiating ( $1$ ) with respect to x, treating y as
constant, wega, $18(x+y-2)+32(x-y)=0\dots .(2)$
Differentiating( $1$ ) with respcctioy.trating xasconstnat.
weget, $18(x+y-2)-32(x-y)=0\dots .(3)$
Solving ( $2$ ) and ( $3$ ), we get the coordinates of the
centre $(1,1)$