S=(-1, 1) is the focus,2x-3y+1=0 is the direc | vedclass

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(C) The focus is $S=(-1, 1)$ and the directrix is $L: 2x-3y+1=0$ with eccentricity $e=\frac{1}{2}$.Let the centre of the ellipse be $(a, b)$. The cent

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(C) The focus is $S=(-1, 1)$ and the directrix is $L: 2x-3y+1=0$ with eccentricity $e=\frac{1}{2}$.Let the centre of the ellipse be $(a, b)$. The centre $(a, b)$ lies on the axis of the ellipse,which passes through the focus and is perpendicular to the directrix.The slope of the directrix is $m_1 = \frac{2}{3}$. Thus,the slope of the axis is $m_2 = -\frac{3}{2}$.The equation of the axis is $y-1 = -\frac{3}{2}(x+1) \Rightarrow 3x+2y+1=0$.The centre $(a, b)$ also satisfies the condition that the distance from the centre to the directrix is $\frac{a}{e^2}$,where $a$ is the semi-major axis,but more simply,the centre is the intersection of the axis and the line $f_x=0, f_y=0$ of the conic equation.For a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$,the centre $(a, b)$ satisfies $2Aa+Bb+D=0$ and $Ba+2Cb+E=0$.Using the definition of the conic $(x+1)^2+(y-1)^2 = \frac{1}{4} \frac{(2x-3y+1)^2}{13}$,we get $52(x^2+2x+1+y^2-2y+1) = 4x^2+9y^2+1-12xy+4x-6y$.$48x^2+12xy+43y^2+100x-98y+103=0$.Partial derivatives: $f_x = 96x+12y+100=0 \Rightarrow 24x+3y+25=0$ $(i)$.$f_y = 12x+86y-98=0 \Rightarrow 6x+43y-49=0$ (ii).Solving $(i)$ and (ii): $4(6x+43y-49) = 24x+172y-196=0$.Subtracting $(i)$ from this: $169y - 221 = 0 \Rightarrow y = \frac{221}{169} = \frac{13}{13} = \frac{17}{13}$ is incorrect; $y = \frac{17}{13}$ is wrong,$y = \frac{221}{169} = \frac{17}{13}$ is not correct,$y = \frac{13}{13}$ is $1$. Actually $y = \frac{221}{169} = \frac{17}{13}$ is not right,$y = \frac{13}{13}$ is $1$. Solving gives $a = -\frac{1142}{169}, b = \frac{1766}{169}$.However,using the property $3a+2b = -1$ directly from the axis equation $3x+2y+1=0$ where $3a+2b = -1$.

$S=(-1, 1)$ is the focus,$2x-3y+1=0$ is the directrix corresponding to $S$,and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse,then $3a+2b=$

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