Step 1: Find the slope of the given chord.
The chord joins $(0,1)$ and $(2,0)$. Its slope is \[ m=\frac{0-1}{2-0}=-\frac12. \] The two tangents are parallel to this, so they also have slope $-\dfrac12$.
Step 2: Write the slope of a tangent to the ellipse.
For $\dfrac{x^2}{4}+y^2=1$ we have $a^2=4$, $b^2=1$. The tangent at $(x_1,y_1)$ has slope $m=-\dfrac{b^2x_1}{a^2y_1}=-\dfrac{x_1}{4y_1}$.
Step 3: Match the slopes.
Set $-\dfrac{x_1}{4y_1}=-\dfrac12$, which gives $x_1=2y_1$.
Step 4: Put this back into the ellipse.
\[ \frac{(2y_1)^2}{4}+y_1^2=1\Rightarrow y_1^2+y_1^2=1\Rightarrow 2y_1^2=1. \] So $y_1=\pm\dfrac{1}{\sqrt2}$ and $x_1=\pm\sqrt2$.
Step 5: Name the two points.
They are $P_1\left(\sqrt2,\dfrac{1}{\sqrt2}\right)$ and $P_2\left(-\sqrt2,-\dfrac{1}{\sqrt2}\right)$, which are opposite ends of a diameter.
Step 6: Find the distance.
\[ d=\sqrt{(2\sqrt2)^2+\left(\tfrac{2}{\sqrt2}\right)^2}=\sqrt{8+2}=\sqrt{10}. \] \[ \boxed{\sqrt{10}} \]