Question

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

Answer 1

Here $y=x^3+2x-4$. Let point of contact be $P(\alpha ,\beta )$ $\therefore \beta ={\alpha }^3+2\alpha -4.....\left(i\right)$

Now $\frac{dy}{dx}=3x^2+2\therefore \frac{dy}{dx}{]}_{atP}=3{\alpha }^2+2=m_T$

As the tangent is perpendicular to a line x + 14y +3 = 0 having slope of $\frac{-1}{14}$.

Therefore, slope of tanget shall be 14 as well. That is, $3{\alpha }^2+2=14\Rightarrow \alpha =2,-2$

put values of $\alpha$ in (i) we get : $\beta =8,-16$.

Hence the equation of tangents : y - 8 = 14(x-2) $\Rightarrow$ 14x - y = 20 and,

y + 16 = 14(x+2) $\Rightarrow$ 14x - y + 12 =0.