Let θ be the acute angle between the tangents to the ellipse x ^ 2...

Given, $\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{1} = 1 . . . \left(1\right)$

${x}^{2} + {y}^{2} = 3 . . . \left(2\right)$

On solving equations \left(1\right) \text{ and } \left(2\right)$, we get</p><p>$\dfrac{{x}^{2}}{9} + \dfrac{3 - {x}^{2}}{1} = 1$</p><p>$⇒ {x}^{2} + 27 - 9{x}^{2} = 9$</p><p>$⇒ 8{x}^{2} = 18$</p><p>$⇒ {x}^{2} = \dfrac{9}{4}$$⇒ x = \pm \dfrac{3}{2}$

$⇒ x = \dfrac{3}{2}$

Since, point of intersection in the first quadrant.

Put, $x = \dfrac{3}{2}$ in equation \left(2\right)$, we get$y = \dfrac{\sqrt{3}}{2}$. </p><p>Hence, the point of intersection of the curves$\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{1} = 1$and${x}^{2} + {y}^{2} = 3$in the first quadrant is \left(\dfrac{3}{2}, \dfrac{\sqrt{3}}{2}\right)$

Now, differentiate the equation \left(1\right)$on both sides with respect to$x$, we get</p><p>${y}^{'} = \dfrac{- x}{9y}$</p><p>$∴$Slope of tangent to the ellipse$\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{1} = 1$at$\left(\dfrac{3}{2}, \dfrac{\sqrt{3}}{2}\right)$is${m}_{1} = - \dfrac{1}{3\sqrt{3}}$</p><p>Now, differentiate the equation$\left(2\right)$on both sides with respect to$x$, we get</p><p>${y}^{'} = \dfrac{- x}{y}$</p><p>$∴$Slope of tangent to the circle${x}^{2} + {y}^{2} = 3$at$\left(\dfrac{3}{2}, \dfrac{\sqrt{3}}{2}\right)$is${m}_{2} = - \sqrt{3}$</p><p>So, if acute angle between both curves is$\theta$then$tan\theta = \left|\dfrac{{m}_{1} - {m}_{2}}{1 + {m}_{1}{m}_{2}}\right| = \left|\dfrac{- \dfrac{1}{3\sqrt{3}} + \sqrt{3}}{1 + \left(- \dfrac{1}{3\sqrt{3}}\right) ( - \sqrt{3} )}\right| = \left(\dfrac{2}{\sqrt{3}}\right)$