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
5.1 Derivation
Let us assume the slope of tangent $PT$ be $m$. Therefore, we can say that slope of normal $PN$ is $ - \frac{1}{m}$.
We can write $$m = {\frac{{dy}}{{dx}}_{(h,k)}}$$
Using point-slope form, equation of tangent $PT$ passing through point $(h,k)$ is $$y - k = m(x - h)$$ Now, to find the co-ordinates where tangent intersects $X$-axis, put $y=0$, we get $$\begin{equation} \begin{aligned} 0 - k = m(x - h) \\ \frac{{ - k}}{m} = x - h \\ x = h - \frac{k}{m} \\\end{aligned} \end{equation} $$ Therefore, Co-ordinates of point $T$ is $\left( {h - \frac{k}{m},0} \right)$.
Similarly, Using point-slope form, equation of normal $PN$ passing through point $(h,k)$ is $$y - k = \frac{{ - 1}}{m}(x - h)$$ Now, to find the co-ordinates where tangent intersects $X$-axis, put $y=0$, we get $$\begin{equation} \begin{aligned} 0 - k = \frac{{ - 1}}{m}(x - h) \\ mk = x - h \\ x = h + mk \\\end{aligned} \end{equation} $$ Therefore, Co-ordinates of point $N$ is $\left( {h + mk,0} \right)$.
Note: As we have all the co-ordinates, we can easily find the length of tangent, subtangent, normal and submormal using distance formulae.
