The center of the ellipse ((x+y-2)^2)/9+((x-y)^2)/16=1 isa) (0,0)b) (1,1)c) (1,0)d) (0,1)

Question

Arun · Answer

Standard equation of ellipse is

x^2÷a^2 + y^2÷b^2 = 1

The given equation represents ellipse but the axes is shifted (changed). The axes are now x+y-2=0 and x-y=0. Solving these equations we get x=y=1 which is centre of ellipse.

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The center of the ellipse ((x+y-2)^2)/9+((x-y)^2)/16=1 isa) (0,0)b) (1,1)c) (1,0)d) (0,1)

The center of the ellipse ((x+y-2)^2)/9+((x-y)^2)/16=1 isa) (0,0)b) (1,1)c) (1,0)d) (0,1)

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The center of the ellipse ((x+y-2)^2)/9+((x-y)^2)/16=1 isa) (0,0)b) (1,1)c) (1,0)d) (0,1)

The center of the ellipse ((x+y-2)^2)/9+((x-y)^2)/16=1 isa) (0,0)b) (1,1)c) (1,0)d) (0,1)

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