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A transversal cuts the same branch of a hyperbola x^2 /a^2 − y^2 /b^2 = 1 in P, P′ and the asymptotes in Q, Q′. Prove that

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A transversal cuts the same branch of a hyperbola x2 /a2 − y2 /b2 = 1 in P, P′ and the asymptotes in Q, Q′. Prove that : 

(i) PQ = P'Q' & (ii) PQ' = P'Q

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TPT PQ = P'Q' and PQ' = P'Q 

b2x2 – a2y2 = a2b2 ; Let the transversal be y = mx + c 

b2x2 – a2 (mx + c)2 = a2b2 

(b2 – a2m2)x2 – 2a2mcx – a2 (c2 + b2) = 0

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