Let the eccentricity of an ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ is reciprocal to that of the hyperbola $$2x^2 - 2y^2 = 1$$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ______.

The hyperbola is $$2x^2 - 2y^2 = 1$$, which can be rewritten as $$\dfrac{x^2}{1/2} - \dfrac{y^2}{1/2} = 1$$.

In this form, $$a_h^2 = b_h^2 = \dfrac{1}{2}$$, so $$c_h^2 = 1$$ and the hyperbola’s eccentricity is $$e_h = \dfrac{c_h}{a_h} = \sqrt{2}$$.

Since the ellipse that meets orthogonally must have eccentricity reciprocal to the hyperbola’s, it follows that $$e = \dfrac{1}{\sqrt{2}}$$.

For an ellipse of the form $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$, the eccentricity satisfies $$e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2} \implies b^2 = \frac{a^2}{2}\,.$$

At a point where this ellipse and the given hyperbola intersect orthogonally, the slope of the ellipse $$y' = -\dfrac{b^2 x}{a^2 y}$$ must be perpendicular to the slope of the hyperbola $$y' = \dfrac{x}{y}$$, which gives $$\left(-\frac{b^2 x}{a^2 y}\right)\left(\frac{x}{y}\right) = -1 \implies \frac{b^2 x^2}{a^2 y^2} = 1 \implies b^2 x^2 = a^2 y^2\,.$$

The hyperbola also satisfies $$y^2 = x^2 - \dfrac{1}{2}$$, and substituting this and $$b^2 = \dfrac{a^2}{2}$$ into $$b^2 x^2 = a^2 y^2$$ yields $$\frac{a^2}{2} x^2 = a^2\left(x^2 - \frac{1}{2}\right) \implies \frac{x^2}{2} = x^2 - \frac{1}{2} \implies x^2 = 1$$ and hence $$y^2 = 1 - \dfrac{1}{2} = \dfrac{1}{2}\,.$$

Since this intersection point lies on the ellipse, substituting $$x^2 = 1$$ and $$y^2 = \tfrac12$$ into $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ gives $$\frac{1}{a^2} + \frac{1/2}{a^2/2} = 1 \implies \frac{1}{a^2} + \frac{1}{a^2} = 1 \implies a^2 = 2$$ and therefore $$b^2 = 1\,.$$

The length of the latus rectum of the ellipse is $$\frac{2b^2}{a} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ so its square is $$(\sqrt{2})^2 = 2\,.$$

Therefore, the answer is $$2$$.