Question

Determine the equation(s) of tangent (s) line to the curve y = 4x³ − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Answer 1

Let (x₁, y₁) be a point on the curve where we need to find the tangent(s).
Slope of the given line = $\frac{-1}{9}$

Since, tangent is perpendicular to the given line,
Slope of the tangent = $\frac{-1}{\left(\frac{-1}{9}\right)}=9$
$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be the point where the tangent is drawn to this curve.} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=4{x_1}^3-3x_1+5 \\ \mathrm{Now},y=4x^3-3x+5 \\ \Rightarrow \frac{dy}{dx}=12x^2-3 \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}12{x_1}^2-3 \\ \text{Given that,} \\ \text{slope of the tangent=slope of the perpendicular line} \\ \Rightarrow 12{x_1}^2-3=9 \\ \Rightarrow 12{x_1}^2=12 \\ \Rightarrow {x_1}^2=1 \\ \Rightarrow x_1=\pm 1 \\ \text{Case-1:}{x}_1=1 \\ {y}_1=4{x_1}^3-3x_1+5=4-3+5=6 \\ \therefore \left(x_1,y_1\right)=\left(1,6\right) \\ \text{Slope of the}\mathrm{tangent}\text{=9} \\ \text{Equation of}\mathrm{tangent}\text{is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-6=9\left(x-1\right) \\ \Rightarrow y-6=9x-9 \\ \Rightarrow 9x-y-3=0 \\ \text{Case-2:}{x}_1=-1 \\ {y}_1=4{x_1}^3-3x_1+5=-4+3+5=4 \\ \therefore \left(x_1,y_1\right)=\left(-1,4\right) \\ \text{Slope of the}\mathrm{tangent}\text{=9} \\ \text{Equation of}\mathrm{tangent}\text{is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-4=9\left(x+1\right) \\ \Rightarrow y-4=9x+9 \\ \Rightarrow 9x-y+13=0 \end{gathered}$$