A tangent to the ellipse x² + 4y² = 4 meets the ellipse x² + 2y² = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x² + 2y² = 6 are at right angles.

Let '$$d$$' be the perpendicular distance from the centre of the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ to the tangent drawn at a point $$P$$ on the ellipse. If $${F_1}$$ and $${F_2}$$ are the two foci of the ellipse, then show that $${\left( {P{F_1} - P{F_2}} \right)^2} = 4{a^2}\left( {1 - {{{b^2}} \over {{d^2}}}} \right)$$.