The locus of the mid-point of the line segment joining the point...

Given,

$\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{2} = 1$

Let coordinate of $D$ \left(\dfrac{2cos\theta + 4}{2}, \dfrac{\sqrt{2}sin\theta + 3}{2}\right) ≡ \left(h, k\right)$</p><p>So,$\dfrac{2h - 4}{2} = cos\theta . . . \left(1\right)$</p><p>And$\dfrac{2k - 3}{\sqrt{2}} = sin\theta . . . \left(2\right)$</p><p>$\text{Equation}{\left(1\right)}^{2} + \text{equation}{\left(2\right)}^{2}$, then we get</p><p>${\left(\dfrac{2h - 4}{2}\right)}^{2} + {\left(\dfrac{2k - 3}{\sqrt{2}}\right)}^{2} = 1 ⇒ \dfrac{{\left(x - 2\right)}^{2}}{1} + \dfrac{{\left(y - \dfrac{3}{2}\right)}^{2}}{\left(\dfrac{1}{2}\right)} = 1$</p><p>$∴$Required eccentricity is$e = \sqrt{1 - \dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}$