An easy one so, Find it if you can.-

It is given that:

$\begin{cases} x^2-y^2 = 5 & ...(1) \\ \dfrac {x^2}{18} + \dfrac {y^2}{8} = 1 & ...(2)\end{cases}$

\(\begin{array} {} Eq.1+8\times Eq.2: & x^2-y^2 + \dfrac {4x^2}{9} + y^2 = 5+8 \\ & \Rightarrow \dfrac {13x^2}{9} = 13\quad \Rightarrow x^2 = 9 \quad \Rightarrow x = \pm 3 \\ & \Rightarrow 9-y^2 = 5 \quad \Rightarrow y^2 = 4 \quad \Rightarrow y = \pm 2 \end{array} \)

Since the two curves are symmetrical to the origin $O(0,0)$ , we need only consider one of the four points $(-3,-2),(-3,2),(3,-2),(3,2)$ for angle $\alpha$ . Let us consider the point $(3,2)$ .

Let us consider the gradient of the two curves at point $(3,2)$ :

$\begin{cases} x^2-y^2 = 5 &\Rightarrow 2x-2y\dfrac{dy}{dx} =0 \\ &\Rightarrow m_1 = \dfrac{dy}{dx} = \dfrac {x}{y} = \dfrac {3}{2} & ...(1a) \\ \dfrac {x^2}{18} + \dfrac {y^2}{8} = 1 &\Rightarrow \dfrac{x}{9}+\dfrac {y}{4} \dfrac{dy}{dx} =0 \\ &\Rightarrow m_2 = \dfrac{dy}{dx} = - \dfrac {4x}{9y} = - \dfrac {12}{18} = - \dfrac{2}{3} & ...(2a)\end{cases}$

We note that $m_1m_2 = \frac {3}{2}\left( -\frac{2}{3} \right) = -1$ , therefore the curves are tangent to each other at the points of intersection and hence $\alpha = 90^\circ$ .

Therefore, $\frac {3}{2} \tan{\frac{\alpha}{6}} = \frac {3}{2} \tan {15^\circ} = \boxed{0.402}$

Its The Direct Application Of The Following Property "If An Ellipse And A Hyperbola Are Confocal(They share common focii) Then they cut each other orthogonally".