Question

Find the equations of the normal to the curve $y=4x^3-3x+5$ which are perpendicular to the line 9x-y +5 =0.

Answer 1

Let the normal to the curve $y=4x^3-3x+5$ be at $P(\alpha .\beta ).\therefore \beta =4{\alpha }^3-3\alpha +5...\left(i\right)$
Now $\frac{dy}{dx}=12x^2-3\therefore m_N=-\frac{1}{12{\alpha }^2-3}$ =Slope of normal at $P(\alpha ,\beta )$
As slope of the given line 9x-y +5 =0 is 9 and this line is perpendicular to the normal so, $(-\frac{1}{12{\alpha }^2-3})\times 9=-1$ i.e., $12{\alpha }^2-3=9\Rightarrow {\alpha }^2=1\therefore \alpha =\pm 1,\beta =6,4$ By(i)
Therefore the points are (1,6), (-1,4).
Equations of Normals are: $y-6=-\frac{1}{9}(x-1)$ i.e., x +9y =55 and $y-4=-\frac{1}{9}(x+1)$ i.e., x+9y =35.