Question
The centre of the ellipse $\frac{(x+y-4)^{2}}{16}+\frac{(x-y)^{2}}{4}=1$ is(a) $(0,0)$(b) $(2,2)$(c) $(1,1)$(d) $(1,2)$
Step 1
The given equation is $\frac{(x+y-4)^{2}}{16}+\frac{(x-y)^{2}}{4}=1$. Now, let's make a substitution to simplify the equation. Let $u = x+y-4$ and $v = x-y$. Then, the equation becomes $\frac{u^2}{16} + \frac{v^2}{4} = 1$. This is now in the standard form of Show more…
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