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

Because the ellipse is centered at the origin and its axes are along the co-ordinate axes, its standard equation can be written as

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$$

where $$a$$ and $$b$$ are the semi-axes and we later decide which one is larger.

The given point $$(4,-1)$$ lies on the ellipse, so its coordinates must satisfy the equation:

$$\frac{4^{2}}{a^{2}}+\frac{(-1)^{2}}{b^{2}}=1.$$

This becomes

$$\frac{16}{a^{2}}+\frac{1}{b^{2}}=1. \quad -(1)$$

The second given point $$(-2,2)$$ also lies on the ellipse, giving

$$\frac{(-2)^{2}}{a^{2}}+\frac{2^{2}}{b^{2}}=1,$$

which simplifies to

$$\frac{4}{a^{2}}+\frac{4}{b^{2}}=1. \quad -(2)$$

To make the algebra clearer, set

$$X=\frac{1}{a^{2}}, \qquad Y=\frac{1}{b^{2}}.$$

Then equations (1) and (2) become

$$16X+Y=1, \quad -(1')$$

$$4X+4Y=1. \quad -(2')$$

From equation (2′) we express $$Y$$ in terms of $$X$$:

$$4X+4Y=1 \;\;\Longrightarrow\;\;4Y=1-4X \;\;\Longrightarrow\;\;Y=\frac{1-4X}{4}.$$

Substituting this value of $$Y$$ into equation (1′):

$$16X+\frac{1-4X}{4}=1.$$

Multiplying through by $$4$$ clears the denominator:

$$4\!\left(16X\right)+\left(1-4X\right)=4.$$

That gives

$$64X+1-4X=4.$$

Combining like terms,

$$60X+1=4 \;\;\Longrightarrow\;\;60X=3 \;\;\Longrightarrow\;\;X=\frac{3}{60}=\frac{1}{20}.$$

Remembering $$X=\dfrac{1}{a^{2}},$$ we have

$$\frac{1}{a^{2}}=\frac{1}{20} \;\;\Longrightarrow\;\;a^{2}=20.$$

Next, compute $$Y$$ using $$Y=\dfrac{1-4X}{4}$$:

$$Y=\frac{1-4\!\left(\frac{1}{20}\right)}{4}=\frac{1-\frac{4}{20}}{4}=\frac{1-\frac{1}{5}}{4}=\frac{\frac{4}{5}}{4}=\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{5}.$$

Since $$Y=\dfrac{1}{b^{2}},$$ it follows that

$$\frac{1}{b^{2}}=\frac{1}{5} \;\;\Longrightarrow\;\;b^{2}=5.$$

We now know $$a^{2}=20$$ and $$b^{2}=5$$, with $$a^{2}>b^{2},$$ so the semi-major axis is $$a$$. The eccentricity $$e$$ of an ellipse with major axis $$a$$ and minor axis $$b$$ is

$$e=\sqrt{1-\frac{b^{2}}{a^{2}}}.$$

Substituting the found values,

$$e=\sqrt{1-\frac{5}{20}}=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}.$$

Hence, the correct answer is Option A.