Find the equation of the line?

**mj.alawami** · November 26th 2009, 02:55 AM

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

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

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

**mathaddict** · November 26th 2009, 03:04 AM

Originally Posted by mj.alawami

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

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

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

HI

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

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

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

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

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

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

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

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