Find the equation of the ellipse whose center | vedclass

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(A) Given: Center $C = (2, -3)$,Focus $S = (3, -3)$,Vertex $A = (4, -3)$.The distance $CA = \sqrt{(4-2)^2 + (-3 - (-3))^2} = \sqrt{2^2 + 0^2} = 2$. Th

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$3x^{2} + 4y^{2} - 12x + 24y + 36 = 0$

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(A) Given: Center $C = (2, -3)$,Focus $S = (3, -3)$,Vertex $A = (4, -3)$.The distance $CA = \sqrt{(4-2)^2 + (-3 - (-3))^2} = \sqrt{2^2 + 0^2} = 2$. Thus,$a = 2$.The distance $CS = \sqrt{(3-2)^2 + (-3 - (-3))^2} = \sqrt{1^2 + 0^2} = 1$. Since $CS = ae$,we have $ae = 1$. Therefore,$e = \frac{1}{a} = \frac{1}{2}$.For an ellipse,the distance from the center to the directrix is $\frac{a}{e}$. Since the major axis is horizontal (along $y = -3$),the directrix is a vertical line $x = h$. The distance from center $(2, -3)$ to the directrix is $\frac{a}{e} = \frac{2}{1/2} = 4$.Since the focus is at $x = 3$ (to the right of the center $x = 2$),the directrix must be on the same side as the vertex relative to the center,at $x = 2 + 4 = 6$.Using the definition of an ellipse,$PS = e \cdot PN$,where $P(x, y)$ is a point on the ellipse and $PN$ is the perpendicular distance to the directrix $x = 6$:$\sqrt{(x-3)^2 + (y+3)^2} = \frac{1}{2} |x-6|$$(x-3)^2 + (y+3)^2 = \frac{1}{4} (x-6)^2$$4(x^2 - 6x + 9 + y^2 + 6y + 9) = x^2 - 12x + 36$$4x^2 - 24x + 36 + 4y^2 + 24y + 36 = x^2 - 12x + 36$$3x^2 + 4y^2 - 12x + 24y + 36 = 0$.

Find the equation of the ellipse whose center is at $(2, -3)$,focus is at $(3, -3)$,and one vertex is at $(4, -3)$.

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