The ordinate of all points on the curve y = f | vedclass

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(D) Given the curve $y = \frac{1}{2\sin^2 x + 3\cos^2 x}$.We can rewrite the denominator as $2\sin^2 x + 3(1 - \sin^2 x) = 3 - \sin^2 x$ or $2(1 - \co

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$1/2$ or $1/3$ according as $n$ is an odd or an even integer.

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(D) Given the curve $y = \frac{1}{2\sin^2 x + 3\cos^2 x}$.We can rewrite the denominator as $2\sin^2 x + 3(1 - \sin^2 x) = 3 - \sin^2 x$ or $2(1 - \cos^2 x) + 3\cos^2 x = 2 + \cos^2 x$.So,$y = \frac{1}{2 + \cos^2 x}$.For the tangent to be horizontal,the derivative $\frac{dy}{dx}$ must be zero.$y' = -\frac{1}{(2 + \cos^2 x)^2} \cdot (2\cos x)(-\sin x) = \frac{2\sin x \cos x}{(2 + \cos^2 x)^2} = \frac{\sin 2x}{(2 + \cos^2 x)^2}$.Setting $y' = 0$ gives $\sin 2x = 0$,which implies $2x = n\pi$,or $x = \frac{n\pi}{2}$ for $n \in \mathbb{Z}$.If $n$ is even,$n = 2k$,then $x = k\pi$,so $\cos^2 x = 1$. Then $y = \frac{1}{2 + 1} = 1/3$.If $n$ is odd,$n = 2k+1$,then $x = k\pi + \frac{\pi}{2}$,so $\cos^2 x = 0$. Then $y = \frac{1}{2 + 0} = 1/2$.Thus,the ordinate is $1/2$ or $1/3$ according as $n$ is an odd or an even integer.

The ordinate of all points on the curve $y = \frac{1}{2\sin^2 x + 3\cos^2 x}$ where the tangent is horizontal,is

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