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# If $x=a\cos 2t,y=b{{\sin }^{2}}t$, then at $(-a,b),\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is equal to(a) $1$ (b) $\dfrac{1}{2}$(c) $0$ (d) $\dfrac{1}{\sqrt{2}}$  Answer Verified
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.
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