Question

The slope of the normal to the curve $y=3x^2$ at the point whose $x$-coordinate $2$ is

• A. $\frac{1}{13}$
• B. $\frac{1}{14}$
• C. $\frac{-1}{12}$
• D. $\frac{1}{12}$

Answer 1

The correct option is C $\frac{-1}{12}$

$y=3x^2$. Differentiating the equation of the curve with respect to $x$.
$\frac{dy}{dx}=6x$
Now
Slope${}_{x=2}=\frac{dy}{dx}{|}_{x=2}$
$=6\left(2\right)$
$=12$.
Hence the slope of the tangent at $x=2$ is $12$. Since the normal is perpendicular to the tangent, therefore, the

Slope${}_{\text{normal}}\times$Slope${}_{\text{tangent}}=-1$ or Slope${}_{\text{normal}}=\frac{-1}{{\text{Slope}}_{\text{tangent}}}=\frac{-1}{12}$