Question

Normal to $y=x^3-3x$ which is parallel to 2x+18y=9.

• A. $x+9y=13$
• B. $2x+18y=9$
• C. $2x-9y=9$
• D. $x+9y=20$

Answer 1

Answer: (D)

$x+9y=20$

$y=x^3-3x\Rightarrow \frac{dy}{dx}=3x^2-3$
$\therefore$ Slope of normals is $\frac{-1}{\frac{dy}{dx}}=\frac{-1}{3x^2-3}$
Since it is parallel to $2x+18y=9$ whose slope is $\frac{-1}{9}$
$\therefore -\frac{1}{3x^2-3}=-\frac{1}{9}\Rightarrow x^2-1=3\Rightarrow x^2=4\Rightarrow x=\pm 2$
Hence, two such points are $(2,2)$ and $(-2,-2)$ slope of the normal at which is $-\frac{1}{9}$
Hence their equations are $y+2=-\frac{1}{9}(x+2)$ and $y+2=-\frac{1}{9}(x+2)$
$\Rightarrow x+9y=20$ and $x+9y=20$