The locus of the mid-point of the line segment joining the point $$(4, 3)$$ and the points on the ellipse $$x^2 + 2y^2 = 4$$ is an ellipse with eccentricity

We need to find the eccentricity of the ellipse formed by the locus of the midpoint of the line segment joining $$(4, 3)$$ and points on the ellipse $$x^2 + 2y^2 = 4$$.

The ellipse $$x^2 + 2y^2 = 4$$ can be written as $$\frac{x^2}{4} + \frac{y^2}{2} = 1$$ so a general point on it is $$(2\cos\theta, \sqrt{2}\sin\theta)$$.

If we denote the midpoint of the segment from $$(4,3)$$ to $$(2\cos\theta,\sqrt{2}\sin\theta)$$ by $$(h,k)$$, then

$$h = \frac{4 + 2\cos\theta}{2} = 2 + \cos\theta$$

$$k = \frac{3 + \sqrt{2}\sin\theta}{2}$$

From these expressions we have $$\cos\theta = h - 2$$ and $$\sin\theta = \frac{2k - 3}{\sqrt{2}}$$.

Using the identity $$\cos^2\theta + \sin^2\theta = 1$$ yields

$$(h - 2)^2 + \frac{(2k - 3)^2}{2} = 1$$.

This can be rewritten in standard form as

$$\frac{(h - 2)^2}{1} + \frac{\left(k - \frac{3}{2}\right)^2}{\frac{1}{2}} = 1$$,

so that $$a^2 = 1$$ and $$b^2 = \frac{1}{2}$$ with $$a^2 > b^2$$.

The eccentricity is then

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

The answer is Option C: $$\dfrac{1}{\sqrt{2}}$$.