A and B are foci of hyperbola 25(x−1)2+(y+1)2=k2(3x+4y+5)2,k>1 and P is a variable point on hyperbola such that △APB is right angled isosceles triangle. If k=tan⁡α,α∈[0,2π] and sum of all values of α is aπ2b (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′=7π4

answer is C.

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Detailed Solution

When AP=2a+2ae⇒AP2=AB2+PB2⇒4a2+a2e2+8a2e=4a2e2+4a2e2⇒e2−2e−1=0⇒e=2+222 or 2−222 reject ⇒e=1±2⇒k=1+2α=6712 and 24712 Sum of all values =7π4a=7 and b=2

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