Question

The equation of normal(s) to the curve $y=x^3-3x$ which is/are 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 $y=x^3-3x\cdots \left(1\right)$
$\Rightarrow \frac{dy}{dx}=3x^2-3$
The normal is parallel to the line $2x+18y=9.$
$\therefore$ slope of the normal $=-\frac{1}{\left(dy/dx\right)}=-\frac{1}{9}$ (slope of the line)
$\Rightarrow \frac{dy}{dx}=9$
$\Rightarrow 3x^2-3=9\Rightarrow x=\pm 2$
From equation $\left(1\right)$
$x=2,y=2$ and
$x=-2,y=-2$
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 $x+9y+20=0$