An angle of intersection of the curves, [x^2 a^2 + y^2 b^2 = 1] and x2 + y2 = ab, a b, i…

Question

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 :

Accepted Answer

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,{x^2} + {y^2} = ab$$

$${{2{x_1}} \over {{a^2}}} + {{2{y_1}y'} \over {{b^2}}} = 0$$

$$ \Rightarrow {y_1}' = {{ - {x_1}} \over {{a^2}}}{{{b^2}} \over {{y_1}}}$$ .... (1)

$$\therefore$$ $2{x_1} + 2{y_1}y' = 0$

$$ \Rightarrow {y_2}' = {{ - {x_1}} \over {{y_1}}}$$ ..... (2)

Here (x1y1) is point of intersection of both curves

$$\therefore$$ $$x_1^2 = {{{a^2}b} \over {a + b}},y_1^2 = {{a{b^2}} \over {a + b}}$$

$$\therefore$$ $$\tan \theta  = \left| {{{{y_1}' - {y_2}'} \over {1 + {y_1}'{y_2}'}}} \right| = \left| {{{{{ - {x_1}{b^2}} \over {{a^2}{y_1}}} + {{{x_1}} \over {{y_1}}}} \over {1 + {{x_1^2{b^2}} \over {{a^2}y_1^2}}}}} \right|$$

$$\tan \theta  = \left| {{{ - {b^2}{x_1}{y_1} + {a^2}{x_1}{y_1}} \over {{a^2}y_1^2 + {b^2}x_1^2}}} \right|$$

$$\tan \theta  = \left| {{{a - b} \over {\sqrt {ab} }}} \right|$$

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