The equation of the ellipse whose centre is ( | vedclass

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(B) Given the centre $(h, k) = (2, -3)$.Since the $y$-coordinates of the centre,focus,and vertex are the same,the major axis is horizontal.The distanc

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$\frac{(x - 2)^2}{4} + \frac{(y + 3)^2}{3} = 1$

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(B) Given the centre $(h, k) = (2, -3)$.Since the $y$-coordinates of the centre,focus,and vertex are the same,the major axis is horizontal.The distance from the centre to the focus is $ae = |3 - 2| = 1$.The distance from the centre to the vertex is $a = |4 - 2| = 2$.Thus,$e = \frac{ae}{a} = \frac{1}{2}$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 2^2(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.The equation of the ellipse is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.Substituting the values,we get $\frac{(x - 2)^2}{4} + \frac{(y + 3)^2}{3} = 1$.

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

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