JEE Class main Answered
Sir pls solve the following.
Asked by rsudipto | 04 Jan, 2019, 07:16: PM
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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