An angle of intersection of the curves, ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ and x² + y² = ab, a > b, is :

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,{x^2} + {y^2} = ab$<br><br>${{2{x_1}} \over {{a^2}}} + {{2{y_1}y'} \over {{b^2}}} = 0$<br><br>$\Rightarrow {y_1}' = {{ - {x_1}} \over {{a^2}}}{{{b^2}} \over {{y_1}}}$ .... (1)<br><br>$\therefore$ $2{x_1} + 2{y_1}y' = 0$<br><br>$\Rightarrow {y_2}' = {{ - {x_1}} \over {{y_1}}}$ ..... (2)<br><br>Here (x₁y₁) is point of intersection of both curves<br><br>$\therefore$ $x_1^2 = {{{a^2}b} \over {a + b}},y_1^2 = {{a{b^2}} \over {a + b}}$<br><br>$\therefore$ $$\tan \theta = \left| {{{{y_1}' - {y_2}'} \over {1 + {y_1}'{y_2}'}}} \right| = \left| {{{{{ - {x_1}{b^2}} \over {{a^2}{y_1}}} + {{{x_1}} \over {{y_1}}}} \over {1 + {{x_1^2{b^2}} \over {{a^2}y_1^2}}}}} \right|$$<br><br>$$\tan \theta = \left| {{{ - {b^2}{x_1}{y_1} + {a^2}{x_1}{y_1}} \over {{a^2}y_1^2 + {b^2}x_1^2}}} \right|$$<br><br>$\tan \theta = \left| {{{a - b} \over {\sqrt {ab} }}} \right|$