If (x^2+y^2 )^2=x y, then d y d x : - Vidyadip

MCQ

If $\left(x^2+y^2\right)^2=x y$, then $\frac{d y}{d x}$ :

A
$\frac{y+4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$
B
$\frac{y-4 x\left(x^2+y^2\right)}{x+4\left(x^2+y^2\right)}$
✓
$\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$
D
$\frac{4 y\left(x^2+y^2\right)-x}{y-4 x\left(x^2+y^2\right)}$

✓

Answer

Correct option: C.

$\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$

(C)Here $\left(x^2+y^2\right)^2=x y$
Differentiating w.r.t. $x$$
\begin{array}{l}
2\left(x^2+y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)=x \frac{d y}{d x}+y \\
\Rightarrow 4\left(x^2+y^2\right) x+4\left(x^2+y^2\right) y \frac{d y}{d x}=x \frac{d y}{d x}+y \\
\Rightarrow \quad\left[4 y\left(x^2+y^2\right)-x\right] \frac{d y}{d x}=y-4 x\left(x^2+y^2\right)
\end{array}
$
Hence, $\frac{d y}{d x}=\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$

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