Derivative as a Rate Measure, Tangents and Normals
4.0 Equation of Normal
4.0 Equation of Normal
As we know that the normal is perpendicular to the tangent. Therefore, $$\begin{equation} \begin{aligned} {\text{slope of tangent }} \times {\text{ slope of normal = }} - 1 \\ {\text{slope of normal = }} - \frac{1}{{{\text{slope of tangent }}}} = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},{y_1}} \right)}}}} \\\end{aligned} \end{equation} $$
The equation of normal passing through a point $\left( {{x_1},{y_1}} \right)$ and slope $m'$ is written in point-slope form as
$$y - {y_1} = m'(x - {x_1})$$ and from above equation, the slope of normal is $$m' = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},{y_1}} \right)}}}}$$ Put it in equation of normal, we get $$\begin{equation} \begin{aligned} y - {y_1} = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},{y_1}} \right)}}}}(x - {x_1}) \\ \\\end{aligned} \end{equation} $$
Example 2. Find the equation of tangent and normal to the curve $2y = 3 - {x^2}$ at $(1,1)$.
Solution: To find the equation of tangent and normal using point-slope form, we need to find the slope and the point at which tangent and normal is drawn is given.
Differentiate the equation of curve with respect to $x$, we get $$\begin{equation} \begin{aligned} 2\left( {\frac{{dy}}{{dx}}} \right) = - 2x \\ \frac{{dy}}{{dx}} = - x \\ \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{(1,1)}} = - 1 \\\end{aligned} \end{equation} $$ Therefore, the equation of tangent at $(1,1)$ is $$\begin{equation} \begin{aligned} y - 1 = {\left( {\frac{{dy}}{{dx}}} \right)_{(1,1)}}(x - 1) \\ y - 1 = - 1(x - 1) \\ y - 1 = - x + 1 \\ x + y = 2 \\\end{aligned} \end{equation} $$ and we know that normal is always perpendicular to the tangent, so $$\begin{equation} \begin{aligned} {\text{slope of tangent}} \times {\text{slope of normal = }} - 1 \\ {\text{slope of normal}} = 1 \\\end{aligned} \end{equation} $$ Equation of normal at $(1,1)$ is $$\begin{equation} \begin{aligned} y - 1 = \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{(1,1)}}}}(x - 1) \\ y - 1 = 1(x - 1) \\ y - 1 = x - 1 \\ y - x = 0 \\\end{aligned} \end{equation} $$
