Correct Answer - Option 4 : x + 12y - 158 = 0
Concept:
The slope of the tangent to the curve = \(\rm \frac {dy}{dx}\)
The slope of normal to the curve = \(\rm \frac{-1}{(\frac {dy}{dx})}\)
Point-slope is the general form: y - y₁ = m(x - x₁), Where m = slope
Calculation:
Here, y = 3x2 + 1
\(\rm \frac {dy}{dx}\) = 6x
\(\rm \frac{dy}{dx}|_{x=2}=12\)
Slope of normal to the curve =\(\rm \frac{-1}{(\frac {dy}{dx})}\) = \(\rm \frac{-1}{12}\)
Equation of normal to curve passing through (2, 13) is y - 13 = \(\rm \frac{-1}{12}\)(x - 2)
⇒ 12y - 156 = -x + 2
⇒ x + 12y - 158 = 0
Hence, option (4) is correct.
