Question

# Tangent at any point on the hyperbola x2a2−y2b2=1 cut the axis at A and B respectively. If the rectangle OAPB (where O is origin) is completed then locus of point P is given by None of these

Solution

## The correct option is A Lets draw the diagram with given hyperbola and its tangent Here APBO forms a rectangle. Since A and B represents point on x And y axis. p≡ (x coordinate of A, y coordinate of B) ≡ (h. k) Then      A≡(h,0)              B≡(0,k) The given hyperbola is, x2a2−y2b2=1 Whose tangent ig given by, y=mx±√a2m2−b2 This passes through (h,0) 0=mh±√a2m2−b2 mh=√a2m2−b2 m2h2=a2m2−b2 m2=b2a2−h2          - - - - - - -(1) Tangent also passes through (0,k) k=√a2m2−b2 k2+b2a2=m2          - - - - - -(2) (1) and (2) ⇒ b2a2−h2=k2+b2a2=m2  a2b2=a2k2+a2b2−h2k2−h2b2 dividing throughout by h2k2  o=a2h2−1−b2k2 i.e.,a2h2−b2k2 Since (h,k) gives point P,the locus can be given as a2x2−b2y2=1

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