The locus of the point on the curve y = sin x | vedclass

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(C) Given the curve $y = \sin x$. The slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \cos x_1$. The equation of the tange

Vedclass · Accepted Answer

$x^2(1 - y^2) = (y - \pi)^2$

Vedclass · Answer

(C) Given the curve $y = \sin x$. The slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \cos x_1$. The equation of the tangent line at $(x_1, y_1)$ is $(y - y_1) = \cos x_1(x - x_1)$. Since this tangent passes through the point $(0, \pi)$,we substitute these coordinates into the equation: $(\pi - y_1) = \cos x_1(0 - x_1) = -x_1 \cos x_1$. Since $y_1 = \sin x_1$,we have $\cos x_1 = \pm \sqrt{1 - \sin^2 x_1} = \pm \sqrt{1 - y_1^2}$. Substituting this into the equation: $(\pi - y_1) = -x_1(\pm \sqrt{1 - y_1^2})$. Squaring both sides,we get $(\pi - y_1)^2 = x_1^2(1 - y_1^2)$. Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $x^2(1 - y^2) = (y - \pi)^2$.

The locus of the point on the curve $y = \sin x$ where the tangent drawn at that point always passes through the point $(0, \pi)$ is

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