Question

Find the equation of the normal to the curve $y=3x^2+1$, which passes through $(2,13)$.

• (A) $x+12y+158=0$
• (B) $x-12y-156=0$
• (C) $12x+y-156=0$
• (D) $x+12y-158=0$

Answer 1

The Correct Option is D

Explanation:
The slope of the tangent to the curve $=\frac{dy}{dx}$The slope of normal to the curve $=\frac{-1}{\left(\frac{dy}{dx}\right)}$Point-slope is the general form: $y-y_1=m(x-x_1)$, Where $m=$ slopeHere, $y=3x^2+1$$\frac{dy}{dx}=6x$${\frac{dy}{dx}|}_{x=2}=12$Slope of normal to the curve $=\frac{-1}{\left(\frac{dy}{dx}\right)}=\frac{-1}{12}$Equation of normal to curve passing through $(2,13)$ is:$y-13=\frac{-1}{12}(x-2)$$\Rightarrow 12y-156=-x+2$$\Rightarrow x+12y-158=0$Hence, the correct option is (D).