If a normal drawn at a point P to the curve y | vedclass

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(A) Given the curve $y = \sin x$. The derivative is $\frac{dy}{dx} = \cos x$. The slope of the normal at point $P(h, k)$ is $m = -\frac{1}{\cos h}$. T

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$x^2=y^2-y^4$

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(A) Given the curve $y = \sin x$. The derivative is $\frac{dy}{dx} = \cos x$. The slope of the normal at point $P(h, k)$ is $m = -\frac{1}{\cos h}$. The equation of the normal passing through the origin $(0, 0)$ is $y - 0 = m(x - 0)$,which simplifies to $y = -\frac{1}{\cos h} x$. Since the point $P(h, k)$ lies on the normal,we have $k = -\frac{h}{\cos h}$,which implies $\cos h = -\frac{h}{k}$. Also,since $P(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$. Using the identity $\sin^2 h + \cos^2 h = 1$,we substitute the expressions: $k^2 + (-\frac{h}{k})^2 = 1$. This simplifies to $k^2 + \frac{h^2}{k^2} = 1$,or $k^4 + h^2 = k^2$. Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $x^2 + y^4 = y^2$,which is $x^2 = y^2 - y^4$.

If a normal drawn at a point $P$ to the curve $y=\sin x$ passes through the origin,then the locus of $P$ is

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