The given curves are: \[ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1 \quad \text{and} \quad x^2 - y^2 = c^2. \] Since these curves intersect at a right angle, the values of \( \frac{dy}{dx} \) at the points of intersection will be of opposite sign and reciprocal of each other. Differentiating the first equation w.r.t. \( x \): \[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{-x b^2}{a^2 y} = \frac{-b^2}{a^2} \frac{x}{y}. \] Differentiating the second equation w.r.t. \( x \): \[ 2x - 2y \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{x}{y}. \] Since both curves intersect at right angles: \[ -\left(\frac{b^2}{a^2} \frac{x}{y}\right) \cdot \frac{x}{y} = -1 \quad \Rightarrow \quad \frac{x}{y} = \frac{a}{b}. \] Now, substitute \( x = ak \) and \( y = bk \) in both equations and eliminate \( k \): For the ellipse: \[ \left(\frac{ak}{a}\right)^2 + \left(\frac{bk}{b}\right)^2 = 1 \quad \Rightarrow \quad k^2 = \frac{1}{2}. \] For the hyperbola: \[ (ak)^2 - (bk)^2 = c^2 \quad \Rightarrow \quad k^2 (a^2 - b^2) = c^2. \] Substituting \( k^2 = \frac{1}{2} \) into the hyperbola equation: \[ a^2 - b^2 = 2c^2. \]