Tangents are drawn from the origin to the curve y = sin x, then their point of contact lie on the curve

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Solution

The correct option is D

The equation of tangent at (x,y) is
Y - y = cos x (X - x)
It pass through (0, 0)
Then, o - y = cos x ( 0 - x)
Or $c o s x = \frac{y}{x}$
Given, sin x = y
Squaring and adding Eqs. (i) and (ii), then
$1 = \frac{y^{2}}{x^{2}} + y^{2} \Rightarrow \frac{1}{y^{2}} - \frac{1}{x^{2}} = 1$

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