Question

If $ x^{3}+y^{3}-3axy=0 $ , then $ \frac{dy}{dx} $ equals

• A. $ \frac{ay-x^{2}}{y^{2}-ax} $
• B. $ \frac{ay-x^{2}}{ay-y^{2}} $
• C. $ \frac{x^{2}+ay}{y^{2}+ax} $
• D. $ \frac{x^{2}+ay}{ax-y^{2}} $

Answer 1

Answer: (A)

Given, $x^{3}+y^{3}-3 a x y=0$
On differentiating w.r.t. $x,$ we get
$3 x^{2}+3 y^{2} \cdot \frac{d y}{d x}-3 a\left(x \frac{d y}{d x}+y\right)=0$
$\Rightarrow 3\left(x^{2}-a y\right)+3 \frac{d y}{d x}\left(y^{2}-a x\right)=0$
$\Rightarrow \,\,\,\,\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}$