Let P(asecθ,btanθ) and Q(asecϕ,btanϕ) where θ+ϕ=π2 be two points on the hyperbola x2a2−y2b2=1. if (h,k) is the point of intersection of the normals at P and Q then k is
A
a2+b2a
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B
−a2+b2a
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C
a2+b2b
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D
−a2+b2b
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Solution
The correct option is D−a2+b2b Equations of the normal at P is ax+bycosecθ=(a2+b2)secθ (i) and the equation of the normal at Q(asecϕ,bsecϕ) is ax+bycosecϕ=(a2+b2)secϕ (ii) Subtracting (ii) from (i) we get y=a2+b2b.secθ−secϕcosecθ−cosecϕ so k=y=a2+b2b.secθ−sec(π/2−θ)cosecθ−cosec(π/2−θ)[∵θ+ϕ=π/2] =a2+b2b.secθ−cosecθcosecθ−secθ=−[a2+b2b] Hence, option 'D' is correct