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:
1084 Upvotes
Verified Answer
The correct answer is:
$-\frac{y}{x}$
(A)
We have,
$x^{2}+y^{2}=t+\frac{1}{t}$$\cdots$ (1) $\quad$ and $\quad x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$...(2)
Squaring (1), we get
$\begin{array}{l}
x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}+2 \Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=x^{4}+y^{4}+2 \text {..(from (2)) } \\
\therefore x^{2} y^{2}=1
\end{array}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& x^{2}\left(2 y \frac{d y}{d x}\right)+y^{2}(2 x)=0 \\
\therefore & \frac{d y}{d x}=\frac{-2 x y^{2}}{2 x^{2} y}=\frac{-y}{x}
\end{aligned}$
We have,
$x^{2}+y^{2}=t+\frac{1}{t}$$\cdots$ (1) $\quad$ and $\quad x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$...(2)
Squaring (1), we get
$\begin{array}{l}
x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}+2 \Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=x^{4}+y^{4}+2 \text {..(from (2)) } \\
\therefore x^{2} y^{2}=1
\end{array}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& x^{2}\left(2 y \frac{d y}{d x}\right)+y^{2}(2 x)=0 \\
\therefore & \frac{d y}{d x}=\frac{-2 x y^{2}}{2 x^{2} y}=\frac{-y}{x}
\end{aligned}$
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.