Thread: Find the equation of the line?

Question:
FInd the equation of the normal lines to the curve $\displaystyle y=x^3-3x $ which is parallel to the line of $\displaystyle 2x+18y-9=0 $

Attempt:
$\displaystyle f'(x)=3x^2-3$
$\displaystyle m=9 $
$\displaystyle x=2 ;y=2 $
$\displaystyle y-2=9(x-2) $

Is my equation correct?
If not please answer in steps
Thank you

Originally Posted by mj.alawami

Question:
FInd the equation of the normal lines to the curve $\displaystyle y=x^3-3x $ which is parallel to the line of $\displaystyle 2x+18y-9=0 $

Attempt:
$\displaystyle f'(x)=3x^2-3$
$\displaystyle m=9 $
$\displaystyle x=2 ;y=2 $
$\displaystyle y-2=9(x-2) $

Is my equation correct?
If not please answer in steps
Thank you

HI


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

This is the gradient of tangent . The gradient of normal would be perpendicular to it .

Gradient of normal , call it M2 is $\displaystyle -\frac{1}{9}$ (do u know y)

$\displaystyle (3x^2-3)(M2)=-1$

$\displaystyle -\frac{1}{9}=-\frac{1}{3x^2-3}$

solving this ... $\displaystyle x=\pm 2$

Put this into the original equation (curve) to find the y's .