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.

\[y = x^2 + 4x - 16\]

\[\text { Since }, \left( x_0 , y_0 \right) \text { lies on curve  . Therefore }\]

\[ y_0 = {x_0}^2 + 4 x_0 - 16 . . . \left( 1 \right)\]

\[\text { Now,} y = x^2 + 4x - 16\]

\[ \Rightarrow \frac{dy}{dx} = 2x + 4\]

\[\text { Slope of tangent} = \left( \frac{dy}{dx} \right)_\left( x_0 , y_0 \right) =2 x_0 +4\]

\[\text { Given that The tangent is parallel to the line So,}\]

\[\text { Slope of tangent=slope of the given line}\]

\[2 x_0 + 4 = 3\]

\[ \Rightarrow 2 x_0 = - 1\]

\[ \Rightarrow x_0 = \frac{- 1}{2}\]

\[\text { From} (1),\]

\[ y_0 = \frac{1}{4} - 2 - 16 = \frac{- 71}{4}\]

\[\text { Now, slope of tangent},m=3\]

\[\left( x_0 , y_0 \right) = \left( \frac{- 1}{2}, \frac{- 71}{4} \right)\]

\[\text { Equation of tangent is }\]

\[y - y_0 = m \left( x - x_0 \right)\]

\[ \Rightarrow y + \frac{71}{4} = 3\left( x + \frac{1}{2} \right)\]

\[ \Rightarrow \frac{4y + 71}{4} = 3\left( \frac{2x + 1}{2} \right)\]

\[ \Rightarrow 4y + 71 = 12x + 6\]

\[ \Rightarrow 12x - 4y - 65 = 0\]