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 = $$\frac{-1}{\left(\frac{-1}{9}\right)}=9$$

$$\text{ Let }(x_1,y_1)\text{ be the point where the tangent is drawn to this curve }.$$

$$\text{ Since, the point lies on the curve }.$$

$$\text{ Hence },y_1=4{x_1}^3-3x_1+5$$

$$\text{ Now },y=4x^3-3x+5$$

$$\Rightarrow \frac{dy}{dx}=12x^2-3$$

$$\text{ Slope of the tangent }={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=12{x_1}^2-3$$

$$\text{ Given that },$$

$$\text{ slope of the tangent = slope of the perpendicular line }$$

$$\Rightarrow 12{x_1}^2-3=9$$

$$\Rightarrow 12{x_1}^2=12$$

$$\Rightarrow {x_1}^2=1$$

$$\Rightarrow x_1=\pm 1$$

$$\text{ Case}-1:x_1=1$$

$$y_1=4{x_1}^3-3x_1+5=4-3+5=6$$

$$\therefore (x_1,y_1)=(1,6)$$

$$\text{ Slope of the tangent}=9$$

$$\text{ Equation of tangent is},$$

$$y-y_1=m(x-x_1)$$

$$\Rightarrow y-6=9(x-1)$$

$$\Rightarrow y-6=9x-9$$

$$\Rightarrow 9x-y-3=0$$

$$\text{ Case }-2:x_1=-1$$

$$y_1=4{x_1}^3-3x_1+5=-4+3+5=4$$

$$\therefore (x_1,y_1)=(-1,4)$$

$$\text{ Slope of the tangent }=9$$

$$\text{ Equation of tangent is },$$

$$y-y_1=m(x-x_1)$$

$$\Rightarrow y-4=9(x+1)$$

$$\Rightarrow y-4=9x+9$$

$$\Rightarrow 9x-y+13=0$$