Question

If $ {{x}^{2}}+{{y}^{2}}=t-\frac{1}{t} $ and $ {{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}, $ then $ \frac{dx}{dy} $ is equal to

• A. $ \frac{1}{{{x}^{2}}{{y}^{3}}} $
• B. $ \frac{1}{x{{y}^{3}}} $
• C. $ \frac{1}{{{x}^{2}}{{y}^{2}}} $
• D. $ \frac{1}{{{x}^{3}}y} $
• E. $ \frac{-1}{{{x}^{3}}y} $

Answer 1

Answer: (D)

Given, $ {{x}^{2}}+{{y}^{2}}=t-\frac{1}{t} $ and $ {{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}} $
$ \Rightarrow $ $ {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\frac{1}{{{t}^{2}}}-2 $
$ \Rightarrow $ $ {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{x}^{2}}+{{y}^{4}}-2 $
$ \Rightarrow $ $ {{x}^{2}}{{y}^{2}}+1=0 $
$ \Rightarrow $ $ {{y}^{2}}=\frac{-1}{{{x}^{2}}} $
On differentiating w.r.t. $ x, $
we get $ 2y\frac{dy}{dx}=\frac{2}{{{x}^{3}}}\Rightarrow \frac{dy}{dx}=\frac{1}{{{x}^{3}}y} $