If the curves, ${{{x^2}} \over a} + {{{y^2}} \over b} = 1$ and ${{{x^2}} \over c} + {{{y^2}} \over d} = 1$ intersect each other at an angle of 90$^\circ$, then which of the following relations is TRUE?

${{{x^2}} \over a} + {{{y^2}} \over b} = 1$ ..........(1)<br><br>Differentiating both sides : <br><br>$${{2x} \over a} + {{2y} \over b}{{dy} \over {dx}} = 0 \Rightarrow {y \over b}{{dy} \over {dx}} = {{ - x} \over a}$$<br><br>$\Rightarrow$ ${{dy} \over {dx}} = {{ - bx} \over {ay}}$ ............(2)<br><br>${{{x^2}} \over c} + {{{y^2}} \over d} = 1$ ........(3)<br><br>Differentiating both sides : ${{dy} \over {dx}} = {{ - dx} \over {cy}}$ ...........(4)<br><br>$${m_1}{m_2} = - 1 \Rightarrow {{ - bx} \over {ay}} \times {{ - dx} \over {cy}} = - 1$$<br><br>$\Rightarrow bd{x^2} = - ac{y^2}$ .............(5)<br><br>$$(1) - (3) \Rightarrow \left( {{1 \over a} - {1 \over c}} \right){x^2} + \left( {{1 \over b} - {1 \over d}} \right){y^2} = 0$$<br><br>$$ \Rightarrow {{c - a} \over {ac}}{x^2} + {{d - b} \over {bd}} \times \left( {{{ - bd} \over {ac}}} \right){x^2} = 0$$ (using 5)<br><br>$\Rightarrow (c - a) - (d - b) = 0$<br><br>$\Rightarrow c - a = d - b$<br><br>$\Rightarrow c - d = a - b$