If ${{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}$ and ${{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}$ , then prove that $\dfrac{dy}{dx}=\dfrac{1}{{{x}^{3}}y}$

Answer Verified Verified
Hint: We are given ${{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}$ squaring this equation on both sides with formula ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$we will get the equation which will be ${{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2$. Also we know that ${{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}$ . From these two equation we can write y in terms of x and then easily differentiate the equation to find $\dfrac{dy}{dx}$.

Complete step by step answer:
Now we are given with two equations.
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
  & {{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 \\
Now substituting the value of ${{x}^{4}}+{{y}^{4}}$ from equation (2) we get
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
  & \Rightarrow 2y\dfrac{dy}{dx}=\dfrac{2}{{{x}^{3}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{y{{x}^{3}}} \\

Hence proved that $\dfrac{dy}{dx}=\dfrac{1}{{{x}^{3}}y}$

Note: While taking the square note that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
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 doesn’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.