Search any question & find its solution
Question:
Answered & Verified by Expert
If $x^2+y^2=t-\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}$ then $\frac{d y}{d x}=$
Options:
Solution:
2036 Upvotes
Verified Answer
The correct answer is:
$\frac{-y}{x}$
Given $\mathrm{x}^2+\mathrm{y}^2=\mathrm{t}-\frac{1}{\mathrm{t}}$
$\begin{aligned}
& \text { and } x^4+y^4=t^2+\frac{1}{t^2} \\
& \Rightarrow x^4+y^4=x^4+y^4+2 x^2 y^2+2 \\
& \Rightarrow x^2 y^2+1=0
\end{aligned}$
Now $\frac{d y}{d x} \Rightarrow 2 x y^2+x^2 2 y y l=0$
$\begin{aligned}
& \Rightarrow y^1=\frac{-2 x y^2}{2 x^2 y} \\
& =\frac{-y}{x}
\end{aligned}$
$\begin{aligned}
& \text { and } x^4+y^4=t^2+\frac{1}{t^2} \\
& \Rightarrow x^4+y^4=x^4+y^4+2 x^2 y^2+2 \\
& \Rightarrow x^2 y^2+1=0
\end{aligned}$
Now $\frac{d y}{d x} \Rightarrow 2 x y^2+x^2 2 y y l=0$
$\begin{aligned}
& \Rightarrow y^1=\frac{-2 x y^2}{2 x^2 y} \\
& =\frac{-y}{x}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.