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Now we are given with two equations.

${{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}....................(1)$

${{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}....................(2)$

Somehow we want to get rid of t and find a relation between x and y. To do this we can easily see that equation (2) is related to square of equation (1).

Let us first take square of equation (1) on both sides

Hence now we get ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( t-\dfrac{1}{t} \right)}^{2}}$

We know that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

Applying this formula we get

\[\begin{align}

& {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+{{\left( \dfrac{1}{t} \right)}^{2}}-2t\left( \dfrac{1}{t} \right) \\

& \Rightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2 \\

\end{align}\]

Now substituting the value of ${{x}^{4}}+{{y}^{4}}$ from equation (2) we get

\[{{t}^{2}}+\dfrac{1}{{{t}^{2}}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2\]

Now let us Subtract ${{t}^{2}}-\dfrac{1}{{{t}^{2}}}$ on both the sides so that we have the equation

\[\Rightarrow 2{{x}^{2}}{{y}^{2}}=-2\]

Dividing the equation by 2 to simplify the equation

\[\Rightarrow {{x}^{2}}{{y}^{2}}=-1\]

Now to differentiate we will separate the variables

\[\Rightarrow {{y}^{2}}=\dfrac{-1}{{{x}^{2}}}\]

$\Rightarrow {{y}^{2}}=-{{x}^{-2}}$

Now let us differentiate on both sides. We know differentiation of ${{x}^{n}}$ is $n{{x}^{n-1}}$ using this we get

$2y\dfrac{dy}{dx}=-(-2){{x}^{-2-1}}$

$\begin{align}

& \Rightarrow 2y\dfrac{dy}{dx}=\dfrac{2}{{{x}^{3}}} \\

& \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{y{{x}^{3}}} \\

\end{align}$

Also while solving we come across equation \[{{x}^{2}}{{y}^{2}}=-1\] this might be confusing as the product of two squares is negative. And hence one can be confused that the solution doesnâ€™t exist but note that this is indeed a curve and hence we can always find a slope that is $\dfrac{dy}{dx}$ for this.