\( \dfrac{3\left(\dfrac{\sqrt{5}+1}{4}\right) + 5\left(\dfrac{\sqrt{5}-1}{4}\right)}{5\left(\dfrac{\sqrt{5}+1}{4}\right) - 3\left(\dfrac{\sqrt{5}-1}{4}\right)} = \dfrac{8\sqrt{5}-2}{2\sqrt{5}+8} \)
\( = \dfrac{4\sqrt{5}-1}{\sqrt{5}+4} \times \dfrac{\sqrt{5}-4}{\sqrt{5}-4} \)
\( = \dfrac{20 - 16\sqrt{5} - \sqrt{5} + 4}{-11} \)
\( = \dfrac{17\sqrt{5}-24}{11} \Rightarrow a = 17, \, b = 27, \, c = 11 \)
\( a + b + c = 52 \)
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
