Find the equation of the normal to the curve y = \(3x^2 + 2\) at point (1, 5).
- 6y - x - 29 = 0
-
6y + x - 31 = 0
- y - 6x - 1 = 0
- y - 6x + 1 = 0
The correct answer is: B
Explanation
y = \(3x^2 + 2\)
\(y^1 = \frac{dy}{dx} = 6x\)
Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:
\(^mtangent = y^1(1) = 6 (1) = 6\)
Slope of the Normal Line \( ^mnorma l= - \frac{1}{^mtangent}\)
\(^mnormal = - \frac{1}{6}\)
\(y−y_1= ^mnormal⋅(x−x_1)\)
=y-5=-\(\frac{1}{6}(x-1)\)
=y-5=-\(\frac{1}{6}x+\frac{1}{6}\)
=y=-\(\frac{1}{6}x+\frac{1}{6}+5\)
Multiply through by 6
=6y=-x+1+30
∴6y+ x - 31=0