Tangent to a Curve

Tangent to the curve at a point is an important concept in calculus. It is useful in understanding the relationship between curves and their slopes. The tangent line to the curve is a straight line that touches a curve at a single point without crossing it at that point. These concepts of Tangents and slopes have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concept of the Tangent to the Curve at a Point. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of nineteen questions have been asked on this topic in JEE Main from 2013 to 2023, including two questions in 2013, one question in 2016, one question in 2017, five questions in 2020, two in 2021, seven in 2022 and one in 2023.

Tangent to the Curve at point:

The tangent to a curve at a point P on it is defined as the limiting position of the secant $P Q$ as the point $Q$ approaches the point $P$ provided such a limiting position exists.

The slope of the tangent to the curve $y=f(x)$ at the point $\left(x_0, y_0\right)$ is given by $\left.\frac{d y}{d x}\right]_{\left(x_0, y_0\right)} \quad\left(=f\left(x_0\right)\right)$. So the equation of the tangent at $\left(x_0, y_0\right)$ to the curve $y=f(x)$ is given by $y-y_0=f^{\prime}\left(x_0\right)\left(x-x_0\right)$.

Normal

The normal to the curve at any point P on it is the straight line which passes through $P$ and is perpendicular to the tangent to the curve at $P$

Slope and Equation of Tangent

Let $P\left(x_0, y_0\right)$ be a point on the continuous curve $y=f(x)$, then the slope of the tangent to the curve at point $P$ is

$
\begin{aligned}
& \left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)} \\
& \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}=\tan \theta=\text { slope of tangent at } P
\end{aligned}
$
Where $\theta$ is the angle which the tangent at $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ makes with the positive direction of the $x$-axis as shown in the figure.

• If the tangent is parallel to $x$-axis then $\theta=0^{\circ}$.

  $
  \begin{aligned}
  & \Rightarrow \quad \tan \theta=0 \\
  & \therefore \quad\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}=0
  \end{aligned}
  $

• If the tangent is perpendicular to $x$-axis then $\Theta=90^{\circ}$

  $
  \begin{aligned}
  & \Rightarrow \quad \tan \theta \rightarrow \infty \quad \text { or } \quad \cot \theta=0 \\
  & \therefore\left(\frac{d x}{d y}\right)_{\left(x_0, y_0\right)}=0
  \end{aligned}
  $

Equation of Tangent

Let the equation of curve be $y=f(x)$ and let point $P\left(x_0, y_0\right)$ lies on this curve.

The slope of the tangent to the curve at a point $P$ is

$
\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}
$
Hence, the equation of the tangent at point $P$ is

$
\left(y-y_0\right)=\left(\frac{d y}{d x}\right)_{\left(x 0, y_0\right)} \cdot\left(x-x_0\right)
$

Tangent from External Point

If a point $Q(a, b)$ does not lie on the curve $y=f(x)$, then the equation of possible tangent to the curve $y=f(x)$ (tangent passing through point $Q$ ( $a$, b)) can be found by first getting the point of contact $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ on the curve.

$P\left(x_0, y_0\right)$ lies on the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$, then

$
y_0=f\left(x_0\right)
$
Also, slope of PQ is

$
\frac{y_0-b}{x_0-a}=\left(\frac{d y}{d x}\right)_{\left(x_0, y_0\right)}
$
By solving the above two equations we get the point of contact point $P$.
Using P we can find the equation of tangent PQ .

Recommended Video Based on Equation of Tangent to the Curve

Solved Examples

Example 1: Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a point on the curve $y=x^2+7 x+2$, nearest to the line, $y=3 x-3$, Then the equation of the normal to the curve at $P$ is:
1) $x+3 y+26=0$
2) $x+3 y-26=0$
3) $x-3 y-11=0$
4) $x-3 y+22=0$

Solution

Let $L$ be the common normal to the parabola $y=x^2+7 x+2$ and line $y=3 x-3$
$\Rightarrow$ slope of tangent of $y=x^2+7 x+2$ at $P=3$ $\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\text {for } \mathrm{P}}=3$

$
\Rightarrow 2 x+7=3 \Rightarrow x=-2 \Rightarrow y=-8
$
So $\mathrm{P}(-2,-8)$
Normal at $\mathrm{P}: \mathrm{x}+3 \mathrm{y}+\mathrm{C}=0$
$\Rightarrow C=26$ ( $P$ satifies the line)
Normal : $x+3 y+26=0$

Example 2: The shortest distance between the line $x-y=1$ and the curve $x^2=2 y$ is:
1) $\frac{1}{2}$
2) $\frac{1}{2 \sqrt{2}}$
3) $\frac{1}{\sqrt{2}}$
4) 0

Solution

The shortest distance between curves is always along the common normal.

the slope of the line

$\frac{d y}{d x}=1=$ slope of the line
$P$ is any point on the parabola, and also tangent pass through point $P$
slope of the tangent to the parabola

$
\begin{aligned}
& 2 x=2 \frac{d y}{d x} \\
& \frac{d y}{d x}=x=1 \\
& \Rightarrow y=\frac{1}{2}
\end{aligned}
$
Point $\mathrm{P}=\left(1, \frac{1}{2}\right)$
$\therefore$ Shortest distance $=\left|\frac{1-\frac{1}{2}-1}{\sqrt{1^2+1^2}}\right|=\frac{1}{2 \sqrt{2}}$

Example 3 : If the tangent to the curve $y=x^3-x^2+x$ at the point $(\mathrm{a}, \mathrm{b})$ is also tangent to the curve $\mathrm{y}=5 \mathrm{x}^2+2 \mathrm{x}-25$ at the point $(2,-1)$, then $|2 \mathrm{a}+9 \mathrm{~b}|$ is equal to $\qquad$
1) 195
2) 48
3) 23
4) 50

Solution

Equation of tangent to $y=5 x^2+2 x-25$ at $(2,-1)$

$
\begin{aligned}
& y^{\prime}=10 x+2=20+2=22 \\
& y+1=22(x-2) \\
& y+1=22 x-44 \\
& y=22 x-45
\end{aligned}
$
This is also tangent to $\mathrm{y}=\mathrm{x}^3-\mathrm{x}^2+\mathrm{x}$ at $(\mathrm{a}, \mathrm{b})$

Finding point of intersection

$
\begin{aligned}
& 22 \mathrm{x}-45=\mathrm{x}^3-\mathrm{x}^2+\mathrm{x} \\
\Rightarrow & \mathrm{x}^3-\mathrm{x}^2-21 \mathrm{x}+45=0 \\
\Rightarrow & (\mathrm{x}-3)^2(\mathrm{x}+5)=0
\end{aligned}
$
Tangent at $\mathrm{x}=3 \Rightarrow \mathrm{a}=3$
Put $\mathrm{x}=3$ in $\mathrm{y}=22 \mathrm{x}-45$

$
\begin{aligned}
& \Rightarrow \mathrm{b}=66-45=21 \\
& \therefore|2 \mathrm{a}+9 \mathrm{~b}|=6+189=195
\end{aligned}
$
Ans:195

Example 4: Let $\mathrm{P}(\mathrm{a}, \mathrm{b})$ be a point on the parabola $\mathrm{y}^2=8 \mathrm{x}$ such that the tangent at P passes through the centre of the circle $x^2+y^2-10 x-14 y+65=0$. Let A be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$. Then the value of $\mathrm{A}+\mathrm{B}$ is equal to
1) 0
2) 25
3) 40
4) 65

Solution

$\mathrm{P}(\mathrm{a}, \mathrm{b})$ is point on $\mathrm{y}^2=8 \mathrm{x}$, such that tangent at P pass through centre of

$
x^2+y^2-10 x-14 y+65=0 \quad \text { i.e }(5,7)
$
Tangent at $\mathrm{P}\left(\mathrm{at}^2, 2 \mathrm{at}\right)$ is ty $=\mathrm{x}+\mathrm{at}^2$
$\mathrm{A}=2 \&$ it pass through $(5,7)$
$7 \mathrm{t}=5+2 \mathrm{t}^2$
$\Rightarrow \mathrm{t}=1, \mathrm{t}=\frac{5}{2}$
$\therefore \mathrm{P}\left(\mathrm{at}^2, 2 \mathrm{at}\right) \Rightarrow(2,4)$ when $\mathrm{t}=1$
$\&\left(\frac{25}{2}, 10\right)$ when $t=\frac{5}{2}$
$\therefore \mathrm{A}=2 \times \frac{25}{2}=25$
$\therefore A+B=65$
Hence, the answer is the option(4).

Example 5: Let M and N be the number of points on the curve $\mathrm{y}^5-9 \mathrm{xy}+2 \mathrm{x}=0$, where the tangents to the curve are parallel to x axis and $y$-axis, respectively. Then the value of $\mathrm{M}+\mathrm{N}$ equals $\qquad$
1) 2
2) 3
3) 4
4) 5

Solution
$
\begin{aligned}
& y^5-9 x y+2 x=0 \\
& 5 y^4 \frac{d y}{d x}-9 x \frac{d y}{d x}-9 y+2=0 \\
& \frac{d y}{d x}\left(5 y^4-9 x\right)=9 y-2 \\
& \frac{d y}{d x}=\frac{9 y-2}{5 y^4-9 x}=0 \text { (for horizontal tangent) } \\
& y=\frac{2}{9} \Rightarrow \text { which does rot satisfy the original equation } \Rightarrow m=0
\end{aligned}
$
Now $5 \mathrm{y}^4-9 \mathrm{x}=0$ (for vertical tangent )

$
\begin{aligned}
& 5 y^4(9 y-2)-9 y^5=0 \\
& y^4[45 y-10-9 y]=0 \\
& y=0 \text { or } 36 y=10 \\
& y=\frac{5}{18} \\
& y=0 \Rightarrow x=0 \& y=\frac{5}{18} \Rightarrow x(0,0)\left(x, \frac{5}{18}\right) \\
& N=2 \\
& M+N=0+2=2
\end{aligned}
$

Hence, the answer is the option(1).

Summary

Tangent, slope, and normal the important in the concept of Calculus. The slope of the tangent at a point on the curve is given by the derivative, and the equation of the tangent can be derived using this slope. These concepts are essential for analyzing and understanding the behavior of curves and surfaces.