If (x/a)2 + (y/b)2 = 1 (a > b) and x2 − y2 = c2 cut orthogonally, then:

Concept: two curves cut orthogonally when their tangents at every common point are perpendicular, which means the product of the two tangent slopes equals −1. For a curve given by an implicit equation, that slope comes from differentiating the equation with respect to x and solving for dy/dx.

Applying this to the given curves:

1. Ellipse (x/a)2 + (y/b)2 = 1: differentiating gives 2x/a2 + (2y/b2)(dy/dx) = 0, so its tangent slope is m1 = −b2x/(a2y).

2. Hyperbola x2 − y2 = c2: differentiating gives 2x − 2y(dy/dx) = 0, so its tangent slope is m2 = x/y.

3. Perpendicular tangents at a common point (x, y) require m1m2 = −1, so (−b2x/(a2y))(x/y) = −1, which gives b2x2 = a2y2, that is y2 = (b2/a2)x2.

4. Substituting y2 = (b2/a2)x2 in the ellipse equation: x2/a2 + x2/a2 = 1, so 2x2/a2 = 1, giving x2 = a2/2 and therefore y2 = b2/2.

5. Substituting these in the hyperbola equation: a2/2 − b2/2 = c2, that is a2 − b2 = 2c2.

Cross-check with numbers: take a2 = 6 and b2 = 2, which gives c2 = 2. The common point then has x2 = 3 and y2 = 1, and both curves are satisfied, since 3/6 + 1/2 = 1 and 3 − 1 = 2. The slopes there are m1 = −(2·√3)/(6·1) = −1/√3 and m2 = √3, whose product is −1, so the intersection is indeed perpendicular.

Hence the ellipse and the rectangular hyperbola cut orthogonally exactly when a2 − b2 = 2c2, and the given condition a > b keeps this difference positive.

Contrasting the other relations that appear here:

• a2 + b2 = 2c2 combines the semi-axes by their sum, whereas the derivation above produces their difference.

• b2 − a2 = 2c2 reverses that difference; with a > b its left side is negative while c2 is positive.

• a2 − b2 = c2 keeps the same difference but drops the factor 2 that comes from x2 = a2/2 and y2 = b2/2.