Question

Find the equation of the tangent line to the curve y = x² + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

Answer 1

Let (x₀, y₀) be the point of intersection of both the curve and the tangent.
$$\begin{gathered} y=x^2+4x-16 \\ \mathrm{Since},\left(x_0,y_0\right)\mathrm{lies}\mathrm{on}\mathrm{curve}.\mathrm{Therefore} \\ {\mathrm{y}}_0={x_0}^2+4x_{\mathit{0}}-16...\left(1\right) \\ \mathrm{Now},y=x^2+4x-16 \\ \Rightarrow \frac{dy}{dx}=2x+4 \\ \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_0,y_0\right)}\text{=2}{\mathit{\text{x}}}_{\mathit{0}}\text{+4} \\ \text{Given that The tangent is parallel to the line So,} \\ \text{Slope of tangent=slope of the given line} \\ 2x_0+4=3 \\ \Rightarrow 2x_0=-1 \\ \Rightarrow x_0=\frac{-1}{2} \\ \text{From (1),} \\ {y}_0=\frac{1}{4}-2-16=\frac{-71}{4} \\ \text{Now, slope of tangent,}\mathit{\text{m}}\text{=3} \\ \left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\left(\frac{-1}{2},\frac{-71}{4}\right) \\ \text{Equation of tangent is} \\ \mathrm{y}-{\mathrm{y}}_0=m\left(\mathrm{x}-{\mathrm{x}}_0\right) \\ \Rightarrow \mathrm{y}+\frac{71}{4}=3\left(\mathrm{x}+\frac{1}{2}\right) \\ \Rightarrow \frac{4\mathrm{y}+71}{4}=3\left(\frac{2\mathrm{x}+1}{2}\right) \\ \Rightarrow 4\mathrm{y}+71=12\mathrm{x}+6 \\ \Rightarrow 12\mathrm{x}-4\mathrm{y}-65=0 \end{gathered}$$