Originally Posted by

**Paymemoney** Hi

I need help on the following question;

Find the equation to the normal to the curve $\displaystyle x^3 + y^3 -2xy + 4x = 8$ at the point (0,2)

I have found the eqaution but i don't understand why the answer is x=0.

$\displaystyle 3x^2 + 3y^2 \frac{dy}{dx} - (2x \frac{dy}{dx} + 2y) +4 = 0$

$\displaystyle \frac{dy}{dx} 3y^2 \frac{dy}{dx} +2y -3x^2$

$\displaystyle \frac{dy}{dx} = \frac{-4-2y-3x^2}{3y^2-2x}$

to find the equation i sub the values (0,2)

which gets: $\displaystyle \frac{-2}{3}$

so equation i get is $\displaystyle y=\frac{3}{2}x + 2$

P.S