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Sir pls solve the following.
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Asked by rsudipto | 04 Jan, 2019, 07:16: PM
answered-by-expert Expert Answer
Let point of contact (h, k).
Tangent line passes through origin. Hence slope m of this tangent line :    m = ( k/h )......................(1)
 
since the point (h,k) is on the curve y = sinx, we have,    k = sin(h)   .................(2)
 
tangent to the curve y = sinx has slope,  that is given by   dy/dx  = cos(x) = cos (h) ................(3)
 
since the slope obtained from eqns.(1) and (3) are same, we have  cos(h) = k/h  .................(4)
 
using eqn.(2) and (4), we write,   sin2(h) + cos2(h) = k2 + (k/h)2 = 1   or  h2k2 = h2 - k2 ...............(5)
 
replacing (h,k) by (x,y), we get the eqn. of curve  x2y2 = x2 - y2   ..........................(6)
 
Eqn.(6) of the curve can be written as,  (1/y2) - (1/x2) = 1
 
Answered by Thiyagarajan K | 07 Jan, 2019, 10:04: AM

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