Question

Find the equation of tangents to the curve $y=x^3+2x-4$, which are perpendicular to line $x+14y+3=0.$

Answer 1

Assuming the point of contact be $(x_1,y_1)$.

As it $(x_1,y_1)$ lies on $y=x^3+2x-4.$

$\therefore y_1=x_1^3+2x_1-4$ ............(1)

Given curve,
$y=x^3+2x-4$

Differentiating the curve w.r.t. x, we get

$\frac{dy}{dx}=3x^2+2$

$\Rightarrow {\begin{array}{c}\frac{dy}{dx}\end{array}]}_{(x_1,y_1)}=3x_1^2+2$ $\left[\text{slope of the tangent at}\right(x_1,y_1)$]

Since the tangent at $(x_1,y_1)$ is perependicular to the line $x+14y+3=0$

Therefore, slope of tangnet at $x_1,y_1\times$ slope of line = -1

$\Rightarrow {\begin{array}{c}\frac{dy}{dx}\end{array}]}_{(x_1,y_1)}\times \left(\frac{-1}{14}\right)=-1$

$\Rightarrow (3x_1^2+2)\left(\frac{-1}{14}\right)=-1$

$\Rightarrow x_1=\pm 2$

Now, when $x_1=2,y_1=(2{)}^3+2(2)-4=8$

When $x_1=-2,y_1=(-2{)}^3+2(-2)-4=-16$

So, the coordinates of the point of contact are (2,8) and (-2,-16).

Equation of tangent for the coordinates (2,8) is:

$y-8=14(x-2)$

$\Rightarrow 14x-y-20=0$

Equation of tangent for the coordinates (-2,-16) is:

$y+16=14(x+2)$

$\Rightarrow 14x-y+12=0$.