An angle of intersection of the curves, x ^ 2 a ^ 2 + y ^ 2 b ^ 2 =...

Given equations are $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1 . . . \left(1\right)$ and ${x}^{2} + {y}^{2} = ab . . . . . \left(2\right)$where $a > b$.

From \left(1\right)$,</p><p>${b}^{2}{x}^{2} + {a}^{2}{y}^{2} = {a}^{2}{b}^{2}$</p><p>Substituting value of${y}^{2}$from \left(2\right)$ in the above equation

\left({b}^{2}{x}^{2} + {a}^{2}\left(ab - {x}^{2}\right)\right) = {a}^{2}{b}^{2}$</p><p>$⇒ \left({b}^{2} - {a}^{2}\right){x}^{2} = {a}^{2}{b}^{2} - {a}^{3}b$<br />$⇒ {x}^{2} = \dfrac{b{a}^{2}\left(b - a\right)}{{b}^{2} - {a}^{2}} = \dfrac{{a}^{2}b}{a + b}$</p><p>$∴ x = \sqrt{\dfrac{{a}^{2}b}{a + b}}$</p><p>Substituting value of${x}^{2}$in \left(2\right)$, we get

${y}^{2} = \dfrac{a{b}^{2}}{a + b}$

$∴ y = \sqrt{\dfrac{a{b}^{2}}{a + b}}$
Point of intersection is \left(\sqrt{\dfrac{{a}^{2}b}{a + b}}, \sqrt{\dfrac{a{b}^{2}}{a + b}}\right)$</p><p>Now, from \left(1\right)$,
$\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

$⇒ \dfrac{2x}{{a}^{2}} + \dfrac{2y}{{b}^{2}}\dfrac{dy}{dx} = 0$
$⇒ \dfrac{dy}{dx} = {m}_{1} = - \dfrac{{b}^{2}x}{{a}^{2}y}$

From \left(2\right)$,<br />${x}^{2} + {y}^{2} = ab$</p><p>$⇒ 2x + 2y \cdot \dfrac{dy}{dx} = 0$</p><p>$⇒ \dfrac{dy}{dx} = {m}_{2} = - \dfrac{x}{y}$</p><p>Let$\theta$be the angle between curve \left(1\right) \text{ and } \left(2\right)$.
$∴ tan\theta = \left|\dfrac{{m}_{1} - {m}_{2}}{1 + {m}_{1}{m}_{2}}\right| = \left|\dfrac{\dfrac{- {b}^{2}x}{{a}^{2}y} + \dfrac{x}{y}}{1 + \dfrac{{b}^{2}{x}^{2}}{{a}^{2}{y}^{2}}}\right| = \left|\dfrac{xy\left({a}^{2} - {b}^{2}\right)}{{b}^{2}{x}^{2} + {a}^{2}{y}^{2}}\right| = \left|\dfrac{xy\left({a}^{2} - {b}^{2}\right)}{{a}^{2}{b}^{2}}\right|$

$∴ {\left(tan\theta\right)}_{\left(\sqrt{\dfrac{{a}^{2}b}{a + b}}, \sqrt{\dfrac{a{b}^{2}}{a + b}}\right)} = \left|\sqrt{\dfrac{{a}^{3}{b}^{3}}{( a + b{)}^{2}}} \cdot \dfrac{\left({a}^{2} - {b}^{2}\right)}{{a}^{2}{b}^{2}}\right| = \left|\dfrac{( a - b )}{\sqrt{ab}}\right|$

Hence, $\theta = \tan^{- 1}\left(\dfrac{a - b}{\sqrt{ab}}\right)$