Find the locus of the points of intersection of two tangents to a hyperbola x2a2−y2b2=1, if sum of their slopes is a constant =λ.
A
(λx2−a2)=2xy
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B
λ(x2−a2)=2xy
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C
(x2−λa2)=2xy
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D
None of these
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Solution
The correct option is Bλ(x2−a2)=2xy Any tangent to hyperbola is y=mx+√a2m2−b2 If it passes through the point (h,k), then (k−mh)2=a2m2−b2 or m2(h2−a2)−2mhk+k2+b2=0 ...(1) Then there will be two tangents passing through (h,k) whose slopes are given by (1) Now m1+m2=λ∴2hkh2−a2=λ ∴ Locus is λ(x2−a2)=2xy