Question

Equation of the normal of curve $y=x^3-2x+4$ at point $(1,3)$ is __________.

• A. $x-y+4=0$
• B. $x-y-4=0$
• C. $x+y-4=0$
• D. $x+y+4=0$

Answer 1

The correct option is C $x+y-4=0$

$y=x^3-2x+4$
$\therefore \frac{dy}{dx}=3x^2-2$
$\therefore {\left(\frac{dy}{dx}\right)}_{\left(1.3\right)}=3\left(1\right)-2=3-2=1$
$\therefore$ slope of tangent $=1$
$\therefore$ slope of normal $=-1$
$\therefore$ We get equation of Normal using
$u-y_1=m(x-x_1)$
$\therefore y-3=-1(x-1)$
$\therefore y-3=-x+1$
$\therefore x+y-3-1=0$
$\therefore$ Required equation of Normal is $x+y-4=0$