Tangent to a Curve

Tangent to a Curve

Komal MiglaniUpdated on 02 Jul 2025, 07:56 PM IST

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.

This Story also Contains

  1. Tangent to the Curve at point:
  2. Slope and Equation of Tangent
  3. Equation of Tangent
  4. Tangent from External Point
  5. Solved Examples Based on Equation of Tangent to the Curve
Tangent to a Curve
Tangent to a Curve

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 Based on Equation of Tangent to the Curve

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).

Frequently Asked Questions (FAQs)

Q: How do tangent lines help in understanding the concept of function approximation?
A:
Tangent lines provide the simplest form of local linear approximation for a function. This concept is fundamental in numerical methods, error analysis, and understanding how complex functions can be approximated by simpler ones over small intervals.
Q: What's the connection between tangent lines and Taylor series expansions?
A:
The first-order Taylor polynomial of a function at a point is precisely the equation of the tangent line at that point. Higher-order terms in the Taylor series provide increasingly accurate polynomial approximations of the function near the point of expansion.
Q: How do tangent lines relate to the concept of linearization in multivariable calculus?
A:
In multivariable calculus, linearization extends the idea of tangent lines to tangent planes or hyperplanes. These linear approximations of multivariable functions near a point are crucial for understanding local behavior and in optimization problems.
Q: What's the significance of tangent lines in understanding function transformations?
A:
When a function is transformed (shifted, stretched, or reflected), its tangent lines undergo corresponding transformations. This helps in visualizing how the function's rate of change is affected by various transformations.
Q: How do tangent lines help in understanding the behavior of piecewise functions?
A:
Tangent lines can help identify if a piecewise function is continuous and/or differentiable at the points where the function definition changes. The existence and continuity of tangent lines at these points indicate the smoothness of the transition between pieces.
Q: What's the relationship between tangent lines and the definition of convex and concave functions?
A:
For a convex function, any tangent line lies entirely below the graph (except at the point of tangency). For a concave function, any tangent line lies entirely above the graph. This property is crucial in optimization and inequality problems.
Q: How do tangent lines relate to the concept of differentials in calculus?
A:
Differentials represent small changes in variables. Geometrically, they correspond to small movements along the tangent line to a curve. The ratio of differentials (dy/dx) represents the slope of this tangent line, connecting to the derivative concept.
Q: How do you determine if a curve has a vertical tangent line?
A:
A curve has a vertical tangent line at points where its derivative is undefined or approaches infinity. This often occurs at cusps or where the function's rate of change becomes extremely large in a vertical direction.
Q: What's the connection between tangent lines and the fundamental theorem of calculus?
A:
The fundamental theorem of calculus relates derivatives to integrals. Geometrically, it shows that the area under a curve (integral) can be found using antiderivatives, which are related to the slopes of tangent lines (derivatives) at each point.
Q: How do tangent lines help in understanding the behavior of trigonometric functions?
A:
Tangent lines to trigonometric functions help visualize their periodic nature and rate of change. For example, the tangent line to sin(x) at x = 0 has a slope of 1, which is key in small angle approximations and Taylor series expansions.