If $$P_1$$ and $$P_2$$ are two points on the ellipse $$\dfrac{x^2}{4} + y^2 = 1$$ at which the tangents are parallel to the chord joining the points $$(0, 1)$$ and $$(2, 0)$$, then the distance between $$P_1$$ and $$P_2$$ is

The given ellipse is $$\dfrac{x^{2}}{4}+y^{2}=1$$ with semi-major axis $$a=2$$ and semi-minor axis $$b=1$$.

The chord joining the points $$(0,1)$$ and $$(2,0)$$ has slope
$$m_{\,\text{chord}}=\dfrac{0-1}{2-0}=-\dfrac12.$$

If a tangent to the ellipse is parallel to this chord, its slope must also be $$-\dfrac12.$$ For the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,$$ the slope of the tangent at any point $$(x_1,y_1)$$ is

$$m=-\dfrac{b^{2}x_1}{a^{2}y_1}.$$

Substituting $$a^{2}=4,\; b^{2}=1$$ and equating the slope to $$-\dfrac12$$ gives

$$-\dfrac{x_1}{4y_1}=-\dfrac12 \quad\Longrightarrow\quad \dfrac{x_1}{4y_1}=\dfrac12 \;\Longrightarrow\; x_1=2y_1.$$

The point also lies on the ellipse, so

$$\dfrac{(2y_1)^{2}}{4}+y_1^{2}=1 \;\Longrightarrow\; \dfrac{4y_1^{2}}{4}+y_1^{2}=1 \;\Longrightarrow\; y_1^{2}+y_1^{2}=1 \;\Longrightarrow\; 2y_1^{2}=1 \;\Longrightarrow\; y_1=\pm\dfrac1{\sqrt2}.$$

Corresponding $$x$$-coordinates are $$x_1=2y_1=\pm\sqrt2.$$ Hence the required points are $$P_1\bigl(\sqrt2,\;\dfrac1{\sqrt2}\bigr),\quad P_2\bigl(-\sqrt2,\;-\dfrac1{\sqrt2}\bigr).$$

The distance between $$P_1$$ and $$P_2$$ is

$$\sqrt{(\sqrt2-(-\sqrt2))^{2}+\left(\dfrac1{\sqrt2}-\left(-\dfrac1{\sqrt2}\right)\right)^{2}} =\sqrt{(2\sqrt2)^{2}+\left(\dfrac{2}{\sqrt2}\right)^{2}} =\sqrt{8+2}= \sqrt{10}.$$

Therefore, the distance between $$P_1$$ and $$P_2$$ is $$\sqrt{10}.$br/> Option D which is: $$$$\sqrt{10}$$$$