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Tangents are drawn from the origin to the curve y=sinx then prove that their point of contact lie on the curve 1/y^2-1/x^2 = 1

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7 months ago

```							Let the tangent from the origin to the curve y = sin x meet the curve again at (x1, y1) Equation of tangent at (x1, y1) is y – y1 = cos(x1)(x – x1) since it passes through the origin, so y1 = x1cos(x1) ...(i) Also, the point (x1, y1) lies on the curve, so y1 = sin (x1) ...(ii) From (i) and (ii), we get, sin(x1) = x1cos(x1) ⇒ x1 = tan(x1) ⇒ x12 = tan2(x1) ⇒ x12 = sec2(x1) – 1 ⇒ x12 = (x12/y12) – 1 ⇒ x12 y12 = x12 – y12 Hence, the locus of (x1, y1) is x2 y2 = x2 – y2
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7 months ago
```							Dear student 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 = sinhslope 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 hand we know sin^2h + cos^2 h =1so locus is k^2 + (k/h)^2 = 1=> h^2k^2 = h^2 - k^2which is  x^2y^2 = x^2 - y^2
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7 months ago
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