If $${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1\left( {a > b} \right)$$      and $${x^2} - {y^2} = {c^2}$$   cut at right angles, then :

Solution :
Let $$\left( {{x_1},\,{y_1}} \right)$$  be their point of intersection then
$$\eqalign{ & x_1^2 - y_1^2 = {c^2}......\left( 1 \right) \cr & \frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} = 1......\left( 2 \right) \cr & \Rightarrow x_1^2\left( {\frac{1}{{{a^2}}} - \frac{1}{{{c^2}}}} \right) + y_1^2\left( {\frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right) = 0......\left( 3 \right) \cr} $$
Now tangents to the curves are $$x{x_1} - y{y_1} = {c^2}$$    and $$\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$$
The tangents are perpendicular, so
$$\frac{{{x_1}}}{{{y_1}}} \times - \frac{{{b^2}}}{{{a^2}}}\frac{{{x_1}}}{{{y_1}}} = - 1 \Rightarrow {b^2}x_1^2 - {a^2}y_1^2 = 0......\left( 4 \right)$$
Eliminating $$x_1^2$$ and $$y_1^2$$ from $$\left( 3 \right)$$ and $$\left( 4 \right),$$ we get,
$$\frac{{{c^2} - {a^2}}}{{{a^2}{b^2}{c^2}}} = \frac{{ - \left( {{b^2} + {c^2}} \right)}}{{{a^2}{b^2}{c^2}}} \Rightarrow {a^2} - {b^2} = 2{c^2}$$