Question

Point of contact tangents to sinx from origin

Answer 1

let the point of contact is h,k  so line passing through origin is k = mh  ...  (1)having slope  m   point also lie on the curve so k = sinh

slope of tangent of curve y = sin x  is   dy / dx =  cos x    which is equal to m     so m = cos h ...(2)  we have m = k/h from (1)      putting in 2 we get  k/h = cos h  

and we know sin2h + cos2 h =1  so locus is k2 + (k/h)2 = 1 ===> h2k2 = h2 - k2                                                         which is  x2y2 = x2 - y2