Question

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

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

Answer 1

Answer: (B)

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