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(A) Given,length of major axis $2a = 6 \Rightarrow a = 3$ and length of minor axis $2b = 2 \Rightarrow b = 1$.The major axis is along the line $x-y+1=

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$(x+y-11)^2+9(x-y+1)^2=18$

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(A) Given,length of major axis $2a = 6 \Rightarrow a = 3$ and length of minor axis $2b = 2 \Rightarrow b = 1$.The major axis is along the line $x-y+1=0$. The minor axis is perpendicular to the major axis and passes through the center $(5,6)$.The slope of the major axis is $m_1 = 1$. Therefore,the slope of the minor axis is $m_2 = -1$.The equation of the minor axis is $y - 6 = -1(x - 5) \Rightarrow x + y - 11 = 0$.The equation of the ellipse with center $(h,k) = (5,6)$ is given by $\frac{(	ext{distance from major axis})^2}{b^2} + \frac{(	ext{distance from minor axis})^2}{a^2} = 1$.Here,the distance from the major axis $x-y+1=0$ is $\frac{|x-y+1|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y+1|}{\sqrt{2}}$.The distance from the minor axis $x+y-11=0$ is $\frac{|x+y-11|}{\sqrt{1^2+1^2}} = \frac{|x+y-11|}{\sqrt{2}}$.Substituting these into the standard form $\frac{d_1^2}{b^2} + \frac{d_2^2}{a^2} = 1$:$\frac{(\frac{x-y+1}{\sqrt{2}})^2}{1^2} + \frac{(\frac{x+y-11}{\sqrt{2}})^2}{3^2} = 1$$\frac{(x-y+1)^2}{2} + \frac{(x+y-11)^2}{18} = 1$Multiplying by $18$:$9(x-y+1)^2 + (x+y-11)^2 = 18$.

An ellipse has $6$ and $2$ as the lengths of its major and minor axes,respectively. If the center is at $(5,6)$ and the major axis is along $x-y+1=0$,then the equation of the ellipse is

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