Maths > Derivative as a Rate Measure, Tangents and Normals > 5.0 Length of tangent, subtangent, normal and subnormal
Derivative as a Rate Measure, Tangents and Normals
1.0 Derivative as a rate of change
2.0 Tangent and Normal
3.0 Equation of Tangent
4.0 Equation of Normal
5.0 Length of tangent, subtangent, normal and subnormal
6.0 Angle of intersection of two curves
5.2 Another method to find the length
As shown in figure, $\theta $ is the angle between $PT$ and $X$-axis and $\theta '$ is the angle between $PN$ and 
$X$-axis. We can say $${\text{slope of line }}PT = m = \tan \theta = {\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}}$$and In $\Delta PST$, $$\begin{equation} \begin{aligned} \tan \theta = \frac{{PS}}{{ST}} = {\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}} \\ \therefore ST = PS{\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}} = {\text{length of sub - tangent}} \\ ST = k{\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}} = {\text{length of sub - tangent}} \\\end{aligned} \end{equation} $$In $\Delta PSN$, $$\begin{equation} \begin{aligned} \tan \theta ' = \frac{{PS}}{{SN}} = - \frac{1}{{\tan \theta }}\quad \left( {\because \tan \theta .\tan \theta ' = - 1} \right) \\ \Rightarrow SN = PS\tan \theta = {\text{length of sub - normal}} \\ SN = k{\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}} = {\text{length of sub - normal}} \\\end{aligned} \end{equation} $$
$X$-axis. We can say $${\text{slope of line }}PT = m = \tan \theta = {\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}}$$and In $\Delta PST$, $$\begin{equation} \begin{aligned} \tan \theta = \frac{{PS}}{{ST}} = {\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}} \\ \therefore ST = PS{\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}} = {\text{length of sub - tangent}} \\ ST = k{\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}} = {\text{length of sub - tangent}} \\\end{aligned} \end{equation} $$In $\Delta PSN$, $$\begin{equation} \begin{aligned} \tan \theta ' = \frac{{PS}}{{SN}} = - \frac{1}{{\tan \theta }}\quad \left( {\because \tan \theta .\tan \theta ' = - 1} \right) \\ \Rightarrow SN = PS\tan \theta = {\text{length of sub - normal}} \\ SN = k{\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}} = {\text{length of sub - normal}} \\\end{aligned} \end{equation} $$Tip to remember: As we know slope of tangent at point $(h,k)$ is ${\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}}$ and slope of normal at point $(h,k)$ is $ - {\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}}$.
We have to remember that in length of subtangent, we use slope of normal and in length of subnormal, we use slope of tangent and multiply that slope with $y$-co-ordinate of point at which tangent/normal is drawn.
Now we have to calculate the length of tangent and length of normal using the above derived results.
In $\Delta PST$, use pythagoras theorem, $$\begin{equation} \begin{aligned} P{T^2} = S{T^2} + P{S^2} \\ {\left( {{\text{length of tangent}}} \right)^2} = {k^2}\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}^2 + {k^2} \\ {\text{ = }}{k^2}\left[ {1 + \left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}^2} \right] \\ \therefore {\text{length of tangent}} = \left| {k\sqrt {\left[ {1 + \left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}^2} \right]} } \right| \\\end{aligned} \end{equation} $$
Similarly, In $\Delta PSN$, use pythagoras theorem, $$\begin{equation} \begin{aligned} P{N^2} = S{N^2} + P{S^2} \\ {\left( {{\text{length of normal}}} \right)^2} = {k^2}\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}^2 + {k^2} \\ {\text{ = }}{k^2}\left[ {1 + \left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}^2} \right] \\ \therefore {\text{length of normal}} = \left| {k\sqrt {\left[ {1 + \left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}^2} \right]} } \right| \\\end{aligned} \end{equation} $$
Example 3. Find the lengths of tangent, subtangent, normal and subnormal to the curve ${y^2} = 4ax$ at $(a{t^2},2at)$.
Solution: From the above explained concept, to find the length of tangent, subtangent, normal and subnormal we need to find the value of $\frac{{dy}}{{dx}}$ at point $(a{t^2},2at)$ and then put the values in the above derived results.
Differentiate the equation of given curve with respect to $x$, we get $$\begin{equation} \begin{aligned} 2y\frac{{dy}}{{dx}} = 4a \\ {\left( {\frac{{dy}}{{dx}}} \right)_{(a{t^2},2at)}} = \frac{{4a}}{{4at}} = \frac{1}{t} \\\end{aligned} \end{equation} $$ Therefore, the length of tangent at $(a{t^2},2at)$ is $$\begin{equation} \begin{aligned} = k\sqrt {1 + \left( {\frac{{dx}}{{dy}}} \right)_{_{(h,k)}}^2} \\ = 2at\sqrt {1 + \left( {\frac{{dx}}{{dy}}} \right)_{_{(a{t^2},2at)}}^2} \\ = 2at\sqrt {1 + {t^2}} \\\end{aligned} \end{equation} $$ Similarly, length of normal at $(a{t^2},2at)$ is $$\begin{equation} \begin{aligned} = k\sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)_{_{(h,k)}}^2} \\ = 2at\sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)_{_{(a{t^2},2at)}}^2} \\ = 2at\sqrt {1 + \frac{1}{{{t^2}}}} \\ = 2a\sqrt {{t^2} + 1} \\\end{aligned} \end{equation} $$ Now, length of subtangent at $(a{t^2},2at)$ is $$\begin{equation} \begin{aligned} = k{\left( {\frac{{dx}}{{dy}}} \right)_{(h,k)}} \\ = 2at \times t = 2a{t^2} \\\end{aligned} \end{equation} $$ and length of subnormal at $(a{t^2},2at)$ is $$\begin{equation} \begin{aligned} = k{\left( {\frac{{dy}}{{dx}}} \right)_{(h,k)}} \\ = 2at \times \frac{1}{t} = 2a \\\end{aligned} \end{equation} $$
Note: Here $(h,k)$ is the point at which the tangent and normal is drawn.
