The eccentricity of an ellipse having centre | vedclass

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(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse

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$\frac{\sqrt{3}}{2}$

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(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(4, -1)$,we have $\frac{16}{a^2} + \frac{1}{b^2} = 1$,which implies $16b^2 + a^2 = a^2b^2$ $(i)$.Since the ellipse passes through $(-2, 2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which implies $4b^2 + 4a^2 = a^2b^2$ $(ii)$.Equating $(i)$ and $(ii)$,we get $16b^2 + a^2 = 4b^2 + 4a^2$.$12b^2 = 3a^2$,so $a^2 = 4b^2$.Substituting $a^2 = 4b^2$ into $(ii)$,we get $4b^2 + 4(4b^2) = (4b^2)b^2$,which simplifies to $20b^2 = 4b^4$,so $b^2 = 5$.Then $a^2 = 4(5) = 20$.Since $b^2 = a^2(1 - e^2)$,we have $5 = 20(1 - e^2)$.$1 - e^2 = \frac{5}{20} = \frac{1}{4}$.$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$.Therefore,$e = \frac{\sqrt{3}}{2}$.

The eccentricity of an ellipse having centre at the origin,axes along the coordinate axes and passing through the points $(4, -1)$ and $(-2, 2)$ is

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