Question

The equation of the normal(s) to $y=x^3-3x$, which is parallel to $2x+18y=9$ is/are

• A. $x+9y=20$
• B. $x+9y+20=0$
• C. $x-9y=20$
• D. $x-9y+20=0$

Answer 1

Answer: (A), (B)

$x+9y+20=0$

The curve is $y=x^3-3x$ $\left(1\right)$

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

The normal is parallel to the line $2x+18y=9$. then the slope of the normal $=-\frac{1}{\left(dy/dx\right)}=-\frac{1}{9}$ (slope of the line)

$\frac{dy}{dx}=9\Rightarrow 3x^2-3=9\Rightarrow x=\pm 2$

From equation $\left(1\right)$, when $x=2$, $y=2$ and when $x=-2$, $y=-2$

Hence the required normals are

$y-2=-\left(\frac{1}{9}\right)(x-2)$ and $y+2=-\left(\frac{1}{9}\right)(x+2)$

$\Rightarrow x+9y=20$ and $x+9y+20=0$