If the values of 'm' for which straight line $y = m x + 2 \sqrt{5}$ touches the hyperbola $16 x^{2} - 9 y^{2} = 144$ are the roots of the equations $x^{2} - (a + b) x - 4 = 0$ . Then the value of $a + b$ is ________.

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Solution

$\begin{matrix} W h e n l i n e y = m x + c t o n c k e s t h e h y p e r b o l a \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 T h e n c^{2} = a^{2} m^{2} - b^{2} g i v e n l i n e y = m x + 2 \sqrt{5} c = 2 \sqrt{5} a n d h y p e r b o l a 16 x^{2} - 9 y^{2} = 144 \Rightarrow \frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} = 1 \Rightarrow \frac{x^{2}}{9} - \frac{9 y^{2}}{16} = 1 \Rightarrow a^{2} = 9, b^{2} = 16 T h e n, c^{2} = a^{2} m^{2} - b^{2} \Rightarrow {(2 \sqrt{5})}^{2} = 9 m^{2} - 16 \Rightarrow 20 + 16 = 9 m^{2} \Rightarrow m^{2} = 4 \Rightarrow m = \pm 2 m i s r o o t s o f e q u a t i o n x^{2} - (a + b) x - 4 = 0 \Rightarrow x^{2} - (s u m o f r o o t s) x + p r o d u c t o f r o o t s = 0 ∴ a + b = + 2 - 2 = 0 \end{matrix}$

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