$A and B are foci of hyperbola 25 [(x - 1)^{2} + (y + 1)^{2}] = k^{2} (3 x + 4 y + 5)^{2}, k > 1 and P is a$ $variable point on hyperbola such that △ A P B is right angled isosceles triangle. If$ $k = \tan α, α \in [0, 2 π] and sum of all values of α is (\frac{aπ}{2 b}) (where a and b are co-prime numbers), then$

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$a = 6$

$b = 3$

$a + b = 9$

$Sum of all values of α = \frac{7 π}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} When AP = 2 a + 2 ae \\ \Rightarrow {AP}^{2} = {AB}^{2} + {PB}^{2} \\ \Rightarrow 4 a^{2} + a^{2} e^{2} + 8 a^{2} e = 4 a^{2} e^{2} + 4 a^{2} e^{2} \\ \Rightarrow e^{2} - 2 e - 1 = 0 \\ \Rightarrow e = \frac{2 + 2 \sqrt{2}}{2} or \frac{2 - 2 \sqrt{2}}{2} reject \\ \Rightarrow e = 1 \pm \sqrt{2} \Rightarrow k = 1 + \sqrt{2} \\ α = 67 \frac{1}{2} and 247 \frac{1}{2} \end{array}$