Tangents are drawn to the curve y = sin x fro | vedclass

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(B) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$. The slope of the tangent at $(h, k)$ is

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$x^2 y^2 = x^2 - y^2$

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(B) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$. The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = \cos x$. At $(h, k)$,the slope is $\cos h$. The equation of the tangent at $(h, k)$ is $y - k = \cos h(x - h)$. Since the tangent passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$: $0 - k = \cos h(0 - h) \Rightarrow -k = -h \cos h \Rightarrow k = h \cos h$. From $k = \sin h$,we have $\cos h = \frac{k}{h}$. Using the identity $\sin^2 h + \cos^2 h = 1$,we substitute $\sin h = k$ and $\cos h = \frac{k}{h}$: $k^2 + \left(\frac{k}{h}\right)^2 = 1 k^2 + \frac{k^2}{h^2} = 1 h^2 k^2 + k^2 = h^2 h^2 - k^2 = h^2 k^2$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 = x^2 y^2$.

Tangents are drawn to the curve $y = \sin x$ from the origin. The locus of the points of contact is

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