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Q.
If tangents are drawn from the origin to the curve $y=\sin x$, then their points of contact lie on the curve
Application of Derivatives
Solution:
The tangent at $\left(x_1, \sin x_1\right)$ is $y-\sin x_1=\cos x_1\left(x-x_1\right)$
It passes through the origin.
$\sin x_1=x_1 \cos x_1=x_1 \sqrt{1-\sin ^2 x_1}$
$y_1{ }^2=\sin ^2 x_1=x_1^2\left(1-y_1^2\right)$
$ \Rightarrow \left(x_1 y_1\right)\left(x_1 y_1\right)$ lies on the curve
$y^2=x^2\left(1-y^2\right)$
$ \Rightarrow x^2-y^2=x^2 y^2$