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Hint: Use parametric equations and chain rule.

As per the given information, \[x=a\cos 2t,y=b{{\sin }^{2}}t\]

For this problem, first we shall find \[\dfrac{dy}{dt}\And \dfrac{dx}{dt}.\]

So, consider, \[y=b{{\sin }^{2}}t\]

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{2}}t)\]

Taking out the constant term, we get

\[\begin{align}

& \dfrac{dy}{dt}=b.\dfrac{d}{dt}({{\sin }^{2}}t) \\

& \dfrac{dy}{dt}=b(2\sin t)\dfrac{d}{dt}(\sin t) \\

\end{align}\]

We know derivative of $sinx$ is $\cos x$ , so we get

\[\dfrac{dy}{dt}=b(2\sin t)(cost)\]

But we know, $\sin 2t=2\sin t\cos t$ , so above equation becomes,

\[\dfrac{dy}{dt}=b\sin 2t........(i)\]

Now consider,\[x=a\cos 2t\]

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{dx}{dt}=\dfrac{d}{dt}(a\cos 2t)\]

Taking out the constant term and derivative of $\cos x$ is $-\sin x$ , so we get

\[\begin{align}

& \dfrac{dx}{dt}=a(-\sin 2t)\dfrac{d}{dt}(2t) \\

& \dfrac{dx}{dt}=-2a.\sin 2t..........(ii) \\

\end{align}\]

Now dividing equations (ii) by (i), we have

\[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{b\sin 2t}{-2a.\sin 2t}\]

Cancelling like terms, we get

\[\dfrac{dy}{dx}=\dfrac{-b}{2a}\]

Now as this is free from the ‘t’ term, we can directly differentiate to find the second derivative.

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-b}{2a} \right)\]

We know derivative of constant term is zero, so we have

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\]

As the second derivative is free of variable terms, so its value is zero at any point.

Hence the correct answer is option (c).

Note: Instead of deriving with respect to ‘t’ and then dividing, we can find the value of ‘t’ from the given values of ‘x’.

We know,

$\cos 2t=1-{{\sin }^{2}}t$

Substituting this in value of ‘x’, we get

\[\begin{align}

& x=a\cos 2t \\

& \Rightarrow x=a(1-{{\sin }^{2}}t) \\

& \Rightarrow \dfrac{x}{a}+1={{\sin }^{2}}t \\

\end{align}\]

Now substituting this value in the value of ‘y’, we get

\[\begin{align}

& y=b{{\sin }^{2}}t \\

& \Rightarrow y=b\left( \dfrac{x}{a}+1 \right) \\

\end{align}\]

Now we can see that this can be differentiated directly with respect to ‘x’.

But this is a complicated process.

As per the given information, \[x=a\cos 2t,y=b{{\sin }^{2}}t\]

For this problem, first we shall find \[\dfrac{dy}{dt}\And \dfrac{dx}{dt}.\]

So, consider, \[y=b{{\sin }^{2}}t\]

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{2}}t)\]

Taking out the constant term, we get

\[\begin{align}

& \dfrac{dy}{dt}=b.\dfrac{d}{dt}({{\sin }^{2}}t) \\

& \dfrac{dy}{dt}=b(2\sin t)\dfrac{d}{dt}(\sin t) \\

\end{align}\]

We know derivative of $sinx$ is $\cos x$ , so we get

\[\dfrac{dy}{dt}=b(2\sin t)(cost)\]

But we know, $\sin 2t=2\sin t\cos t$ , so above equation becomes,

\[\dfrac{dy}{dt}=b\sin 2t........(i)\]

Now consider,\[x=a\cos 2t\]

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{dx}{dt}=\dfrac{d}{dt}(a\cos 2t)\]

Taking out the constant term and derivative of $\cos x$ is $-\sin x$ , so we get

\[\begin{align}

& \dfrac{dx}{dt}=a(-\sin 2t)\dfrac{d}{dt}(2t) \\

& \dfrac{dx}{dt}=-2a.\sin 2t..........(ii) \\

\end{align}\]

Now dividing equations (ii) by (i), we have

\[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{b\sin 2t}{-2a.\sin 2t}\]

Cancelling like terms, we get

\[\dfrac{dy}{dx}=\dfrac{-b}{2a}\]

Now as this is free from the ‘t’ term, we can directly differentiate to find the second derivative.

Now take differentiation on both sides with respect to ‘t’, we get

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-b}{2a} \right)\]

We know derivative of constant term is zero, so we have

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\]

As the second derivative is free of variable terms, so its value is zero at any point.

Hence the correct answer is option (c).

Note: Instead of deriving with respect to ‘t’ and then dividing, we can find the value of ‘t’ from the given values of ‘x’.

We know,

$\cos 2t=1-{{\sin }^{2}}t$

Substituting this in value of ‘x’, we get

\[\begin{align}

& x=a\cos 2t \\

& \Rightarrow x=a(1-{{\sin }^{2}}t) \\

& \Rightarrow \dfrac{x}{a}+1={{\sin }^{2}}t \\

\end{align}\]

Now substituting this value in the value of ‘y’, we get

\[\begin{align}

& y=b{{\sin }^{2}}t \\

& \Rightarrow y=b\left( \dfrac{x}{a}+1 \right) \\

\end{align}\]

Now we can see that this can be differentiated directly with respect to ‘x’.

But this is a complicated process.

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