Length of Tangent, Subtangent, Normal and Subnormal

Length of Tangent, Normal, Subtangent and Subnormal

Length of Tangent:

The length of the portion lying between the point of tangency i.e. the point on the curve from which a tangent is drawn and the point where the tangent meets the $x$-axis. Here point of tangency is $P\left(x_0, y_0\right)$
In the figure, the length of segment PT is the length of the tangent.
In $\triangle \mathrm{PTS}$

$
\begin{aligned}
\mathrm{PT} & =|y \cdot \csc \theta|=|y| \sqrt{1+\cot ^2 \theta} \\
& =|\mathrm{y}| \sqrt{1+\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)_{\left(\mathrm{x} 0, \mathrm{y}_0\right)}}
\end{aligned}
$

Length of Normal:

A segment of normal PN is called length of Normal.
In $\triangle P S N$

$
\begin{aligned}
\mathrm{PN} & =\left|y \cdot \csc \left(90^{\circ}-\theta\right)\right|=|y \cdot \sec \theta| \\
& =|\mathrm{y}| \sqrt{1+\tan ^2 \theta}=|\mathrm{y}| \sqrt{1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(\mathrm{x} 0, \mathrm{y} 0)}}
\end{aligned}
$

Length of Subtangent:

The projection of the segment PT along the x-axis is called the length of the subtangent. In the figure, ST is the length of the subtangent.

In ΔPST

$\begin{aligned} \mathrm{ST} & =|y \cdot \cot \theta|=\left|\frac{y}{\tan \theta}\right| \\ & =\left|\mathrm{y} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}\right|\end{aligned}$

Length of Subnormal:

$
\begin{aligned}
&\text { The projection of the segment PN along the } \mathrm{x} \text {-axis is called the length of the subnormal. In the figure, } \mathrm{SN} \text { is the length of }\\
&\begin{aligned}
& \operatorname{In} \triangle \mathrm{PSN} \\
& \mathrm{SN}=\left|y \cdot \cot \left(90^{\circ}-\theta\right)\right|=|y \cdot \tan \theta| \\
&=\left|\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dv}}\right|
\end{aligned}
\end{aligned}
$

Solved Examples Based on Length of tangents, normal, subtangent, and subnormal:

Example 1: If the Rolle's theorem holds for the function $f(x)=2 x^3+a x^2+b x$ in the interval $[-1,1]$ for the point $c=\frac{1}{2}$ then the value of $2 a+b$ is :
[JEE Main 2014]
1) -1
2) 1
3) 2
4) -2

Solution
As we have learned
Rolle's Theorems - $\square$
Let $f(x)$ be a function of $x$ subject to the following conditions.
1. $\mathrm{f}(\mathrm{x})$ is continuous function of $x: x \in[a, b]$
2. $\mathrm{F}(\mathrm{x})$ is exists for every point: $x \epsilon[a, b]$
3. $f(a)=f(b)$ then $f^{\prime}(c)=0$ such that $a<c<b$.

Geometrical interpretation of Rolle's theorem -

Let f(x) be a function defined on [a, b] such that the curve y = f(x) is continuous between points {a, f(a)} and {b, f(b)} at every points on the curve encept at the end point it is possible to draw a unique tangent and ordinates at x = a and x = b are equal f(a) = f(b).

$
\begin{aligned}
&\text { - wherein, We have }\\
&\begin{aligned}
& f^{\prime}(1 / 2)=\frac{f(1)-f(-1)}{2}=0 \\
\Rightarrow & \left.\left(6 x^2+2 a x+b\right)\right|_{x=1 / 2} \\
& \frac{2+a+b-(-2+a-b)}{2}=0 \\
\Rightarrow & 3 / 2+a+b=\frac{4+2 b}{2}=0 \\
\Rightarrow & 3+2 a+2 b=4+2 b=0 \\
\Rightarrow & a=1 / 2 \text { and } b=-2 \\
\therefore & 2 a+b=-1
\end{aligned}
\end{aligned}
$

Example 2: If the tangent to the curve $y=x^3$ at the point $P\left(t, t^3\right)$ meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio $1: 2$ is:
[JEE Main 2021]
1) $-2 t^3$
2) $-t^3$
3) 0
4) $2 t^3$

Solution
Equation of tangent at $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^3\right)$

$
\left(y-t^3\right)=3 t^2(x-t)
$

now solve the above equation with

$
y=x^3
$

$
\begin{aligned}
&\begin{aligned}
& \text { By }(1) \&(2) \\
& x^3-t^3=3 t^2(x-t) \\
& \mathrm{x}^2+\mathrm{xt}+\mathrm{t}^2=3 \mathrm{t}^2 \\
& \mathrm{x}^2+\mathrm{xt}-2 \mathrm{t}^2=0 \\
& (x-t)(x+2 t)=0 \\
& \Rightarrow x=-2 t \Rightarrow Q\left(-2 t,-8 t^3\right)
\end{aligned}\\
&\text { Ordinate of required point }\\
&=\frac{2 t^3+\left(-8 t^3\right)}{3}=-2 t^3
\end{aligned}
$

Hence, the answer is the option (1).

Example 3: The shortest distance between the line $\mathrm{y}=\mathrm{x}$ and the curve $y^2=x-2$ is.
[JEE Main 2015]
1) $\frac{7}{4 \sqrt{2}}$
2) $\frac{7}{2 \sqrt{2}}$
3) $\frac{7}{4 \sqrt{3}}$
4) $\frac{5}{4 \sqrt{2}}$

Solution

Line $\mathrm{y}=\mathrm{x}$
Eq. of tangent to $\mathrm{y}^2=\mathrm{x}-2$
$y^2=x-2$
$2 \mathrm{yy}^{\prime}=1$
$\mathrm{y}^{\prime}=\frac{1}{2 \mathrm{y}}=$ slope
Tangent at $P$ is parallel to the line $x=y$
so, slope should be equal

$
\mathrm{y}^{\prime}=1=\frac{1}{2 \mathrm{y}} \Rightarrow \mathrm{y}=\frac{1}{2}
$

put the value of $y$ in the curve $y^2=x-2$

$
\left(\frac{1}{2}\right)^2=x-2 \Rightarrow x=\frac{9}{4}
$

so, $\mathrm{P}=\left(\frac{9}{4}, \frac{1}{2}\right)$
Perpendicular distance from the point $P$ to the line $y=x$

$
\left|\frac{\left(\frac{9}{4}-\frac{1}{2}\right)}{\sqrt{1^2+1^2}}\right|=\frac{7}{4 \sqrt{2}}
$
Hence, the answer is option (1).

Example 4: If Rolle's theorem holds for the function $f(x)=2 x^3+b x^2+c x, x \in[-1,1]$ at the point $x=\frac{1}{2}$, then $2 b+c$ equals:
[JEE Main 2019]
1) -1
2) -2
3) -3
4) -4

Solution
Rolle's Theorems
Let $f(x)$ be a function with the following properties
1. $f(x)$ is a continuous function in $[a, b]$
2. $\underline{\underline{f}}^{\prime}(\mathrm{x})$ exists for every point in (a,b)
3. $f(a)=f(b)$

Then there is at least one c lying in $(a, b)$ such that $\underline{f}(c)=0$
Now,

$
f(x)=2 x^3+b x^2+c x
$

It is continuous and differentiable in any interval as it is a polynomial, hence it is continuous in $[-1,1]$ and differentiable in $(-1,1)$
Now $f(-1)=f(1)$
where, $f(1)=2+b+c$ and $f(-1)=-2+b-c$

$
\begin{aligned}
& \Rightarrow b+c+2=b-2-c \\
& \Rightarrow c+2=0 \\
& \therefore c=-2
\end{aligned}
$
Also

$
f^{\prime}(x)=6 x^2+2 b x+c
$
As Rolle's Theorem is satisfied at $x=1 / 2$, hence $f^{\prime}(1 / 2)=0$

$
\begin{aligned}
& 0=6\left(\frac{1}{2}\right)^2+b \times 2 \times \frac{1}{2}-2 \\
& 0=\frac{6}{4}+b-2 \\
& \mathrm{~b}=1 / 2
\end{aligned}
$
So, $2 b+c=-1$

Hence the answer is the option (1)

Example 5 : Length of normal drawn to the curve $x y=16$ at its point $(4,4)$ equals?
1) $6 \sqrt{2}$
2) $5 \sqrt{2}$
3) $4 \sqrt{2}$
4) $3 \sqrt{2}$

Solution
As we know, the length of Normal $=y_o \sqrt{1+\left(y^{\prime}\right)^2}$
Here the curve is

$
\begin{aligned}
& x y=16 \\
& \Rightarrow y=\frac{16}{x} \\
& \Rightarrow y^{\prime}=-16 / x^2 \\
& \Rightarrow y^{\prime} \text { at }(4,4)=-1 \\
& \therefore \text { Length of normal }=4 \sqrt{1+1}=4 \sqrt{2}
\end{aligned}
$

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NCERT Resources

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Length Of Tangent And Normal And Subtangent And Subnormal- Practice Question MCQ

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