Hello mj.alawami Originally Posted by
mj.alawami So From what i understood?
$\displaystyle 4x^3=\frac{1}{2} $
$\displaystyle x=\frac{1}{2} $
$\displaystyle y-\frac{1}{16}=\frac{1}{2}(x-\frac{1}{2} $
$\displaystyle x-2y+\frac{3}{8} $
Is my answer correct??
If not please answer in steps
Your answer is partly correct. You have the right values of $\displaystyle x$ and $\displaystyle y$, but the wrong gradient. The gradient of the normal is $\displaystyle -2$, so the equation is:
$\displaystyle y-\frac{1}{16}=-2\left(x - \frac12\right)$
i.e. $\displaystyle 16y-1=-32x + 16$
i.e. $\displaystyle 32x+16y-17 =0$
Grandad