Tangent and Normal

Imagine you're driving along a winding mountain road. The road represents a curve, and your car's path at any given moment is like the tangent to that curve, touching it at just one point and showing the direction you're heading. Now, if you were to suddenly stop and look directly to your left or right, you'd be looking along the normal, a line perfectly perpendicular to your path at that point.

In Mathematical terms, tangents and normal are fascinating lines that help us understand curves better. The tangent line touches the curve at one point and shares the same direction as the curve at that point. The normal line, on the other hand, stands at a right angle to the tangent, like a trusty sidekick always ready to provide a different perspective.

In this article, we'll explore how to utilize differentiation to determine the equations of tangent and normal lines to a curve at specified points.

1.0What are Tangents and Normal?

Tangents and normal are geometric concepts frequently encountered in calculus and geometry. 

The tangent to the curve at ‘p’ is the line through P whose slope is the limit of the slopes of secants as Q → P from either side.

In short : A tangent to a curve is the limiting case of secants.

Geometrical Interpretation of Derivative

Let y = f(x) is a function then $\frac{dy}{dx}$ is slope of the tangent i.e. ${\left(\frac{dy}{dx}\right)}_{(x,y)}=\tan \theta$, where θ is the angle made by the tangent with the positive direction of x-axis.  

On the other hand, a normal to a curve at a given point is a straight line that is perpendicular to the tangent at that point, intersecting the curve. It represents the direction perpendicular to the curve at that specific location.

2.0Slopes of the Tangent and the Normal

Slope of the Tangent

Let y = f(x) be a curve, and let P (x₁, y₁) be a point on it. Then, ${\left(\frac{dy}{dx}\right)}_{p}is$ the slope of the tangent to the curve y = f (x) at point P. i.e.; ${\left(\frac{dy}{dx}\right)}_p=\tan \theta$ = slope of the tangent at P, where θ is the angle which the tangent at P(x₁, y₁) makes with the positive direction of x-axis.

Slope of the Normal

The normal to a curve at P (x₁, y₁) is a line perpendicular to the tangent at P and passing through P. Slope of the normal at 

$P=-\frac{1}{\text{ slope of tangent at }P}=-\frac{1}{{\left(\frac{dy}{dx}\right)}_P}=-{\left(\frac{dx}{dy}\right)}_P$

If normal makes an angle of θ with positive direction of x-axis, then 

$-\frac{dx}{dy}=\tan \theta \text{ or }\frac{dy}{dx}=-\cot \theta$

3.0Equation of Tangent and Normal to the Curve at a Point

1. Equation of the tangent at P (x₁, y₁) is:

$y-y_1={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\left(x-x_1\right)$where P is called point of contact 

2. Equation of normal at P (x₁, y₁) is:  

$\left(y-y_1\right)=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}}\left(x-x_1\right).$where P is called the foot of normal.

4.0Tangent and Normal Formula

Tangent Line Formula

The equation of the tangent line y = mx + c to a curve at a point (x₁, y₁) is given by:

$y-y_1=m\left(x-x_1\right)$

where m is the slope of the tangent line, which is obtained by differentiating the function representing the curve and evaluating it at x₁.  

Normal Line Formula

The equation of the normal line y = mx + c to a curve at a point (x₁, y₁) is given by:

$y-y_1=\left(-\frac{1}{m}\right)\left(x-x_1\right)$

where m is the slope of the tangent line, obtained in the same manner as for the tangent line.

In both cases, (x₁, y₁) is the given point on the curve where the tangent or normal line is to be drawn.

5.0Properties of Tangents and Normal

1. Tangent Parallel to X-axis:

• If θ = 0 then Slope = 0

2. Tangent Parallel to Y-axis:

• If $\theta =\frac{\pi }{2}$ then Slope undefined (vertical tangent)

3. Tangent Makes Equal Angles with Axes:

• If $\theta =\frac{\pi }{4}\text{ or }\frac{3\pi }{4}$ then tan θ = ±1

4. Tangent Equally Inclined to Axes:

• If  $\theta =\frac{\pi }{4}\text{ or }\frac{3\pi }{4}$ then tan θ = ±1

5. Tangent with Equal Non-Zero Intercepts:

• Tangent has symmetric intercepts on the axes.
• $\theta =\frac{3\pi }{4}$ then tan θ = –1

Common Parametric Coordinates

There are many curves out there that we can’t even write down as a single equation in terms of only x and y. So, to deal with some of these problems we introduce parametric equations.

Instead of defining y in terms of x or x in terms of y, we define both x and y in terms of a third variable called a parameter as follows,

x = f(t), y = g(t)

(x,y) = (f(t),g(t))

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$

6.0Angle of Intersection of two Curves

The angle of intersection of two curves at a point of intersection (say at P) is defined as the angle between the two tangents/normal to the curve at their point of intersection.

Orthogonal Curves:

If the angle of intersection of two curves at every point of intersection is 90°, then curves are said to be orthogonal, i.e. $\left(\frac{d{y}_1}{dx}\right)\left(\frac{d{y}_2}{dx}\right)=-1$ at every point of intersection.

7.0Application of Tangents and Normal in Real Life

1. Engineering and Construction:

• Road/Railway Design: Smooth transitions between curves and straights.
• Bridge Stability: Analyzing stress and load on arches and beams.

2. Physics:

• Motion Analysis: Determining velocity and analyzing perpendicular forces.
• Optics: Measuring angles of incidence and reflection.

3. Astronomy:

• Orbit Calculations: Studying trajectories and velocities.
• Telescope Alignment: Accurate positioning for celestial observations.

4. Robotics:

• Path Planning: Smooth paths and collision avoidance.
• Manipulator Control: Directing robotic arms.

5. Computer Graphics:

• Rendering: Creating realistic curves and surfaces.
• Animation: Ensuring smooth movements.

8.0Solved Examples of Tangents and Normal

Example 1: Angle between the tangents to the curve y = x² – 5x + 6 at the points (2, 0) and (3, 0) is-

(A) $\frac{\pi }{2}$ (B) $\frac{\pi }{6}$ (C) $\frac{\pi }{4}$ (D) $\frac{\pi }{3}$

Ans. (A)

Solution: Angle between the tangents to the curve y = x² – 5x + 6 is $\frac{dy}{dx}=2x-5$

${\left(\frac{dy}{dx}\right)}_{(2,0)=-1}{\left(\frac{dy}{dx}\right)}_{(3,0)}=1$

$\iff \text{ Angle }=\frac{\pi }{2}$

Example 2: The equation of the tangent to the curve $y=x+\frac{4}{x^2}$, that is parallel to the x-axis, is: -

(A) y = 0 (B) y = 1 (C) y = 2 (D) y = 3 

Ans. (D)

Solution: $y=x+\frac{4}{x^2}$

$\frac{dy}{dx}=1-\frac{8}{x^3}$

Equation of tangent is parallel to the x-axis.

$\therefore \frac{dy}{dx}=0\Rightarrow 1-\frac{8}{x^3}$=0 ⇒ x³ = 8 ⇒ x = 2

At, x = 2, y=2+$\frac{4}{4}$=3 ⇒ y₁ = 3

∴ point is (2, 3)

equation of tangent is: y – y₁ = 0(x – x₁)

y = 3

Example 3: The curve y – e^xy + x = 0 has a horizontal tangent at

(A) (1, 1) (B) (0, 1) (C) (1, 0) (D) no point

Ans. (B) 

Solution: y– e^xy + x = 0 

Differentiating w.r.t. to x

$\frac{dy}{dx}-e^{xy}\left(\frac{dy}{dx}\cdot x+y\right)+1=0$

$\frac{dy}{dx}=0$

1 – ye^xy=0

ye^xy = 1  ⇒ x = 0, y = 1

Point is (0, 1)

Example 4: If the tangent at a point P, with parameter t, on the curve x = 4t² + 3, y = 8t³ – 1, t ∈ R, meets the curve again at a point Q, then the coordinates of Q are:

(A) (t² + 3, t³ – 1) (B) (t² + 3, –t³–1)

(C) (16t² + 3, – 64t³ – 1) (D) (4t² + 3, –8t³ – 1) 

Ans. (B)

Solution:

x=4t² + 3; y = 8t³–1

Slope of tangent $\frac{dy}{dx}=3t$

Also : $P\left(x_1{y}_1\right)=\left(4t_1^2+3,8t_1^3-1\right)$

∴ Slope = 3t₁

$\mathrm{Q}\left({\mathrm{x}}_2{\mathrm{y}}_2\right)=\left(4t_2^2+3,8t_2^3-1\right)$

$\frac{\left(8t_1^3-1\right)-\left(8t_2^3-1\right)}{\left(4t_1^2+3\right)-\left(4t_2^2+3\right)}=3t_1$

$\frac{2\left(t_1^2+t_1{t}_2+t_2^2\right)}{t_1+t_2}=3t_1$

${\mathrm{t}}_1^2+{\mathrm{t}}_1{\mathrm{t}}_2-2{\mathrm{t}}_2^2=0$

${\mathrm{t}}_1=-2{\mathrm{t}}_2\text{ OR }{t}_2=-\frac{t_1}{2}$

∴ option (B) is correct.

Example 5: The slope of the tangent to the curve x = 3t² +1, y = t³–1 at x =1 is

(A) 0 (B) $\frac{1}{2}$ (C) ∞ (D) – 2

Ans. (A)

Solution:

x = 3t² +1, y = t³–1

$\therefore \frac{dx}{dt}=6t,\frac{dy}{dt}=3t^2$

Now $\frac{dy}{dx}=\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)=\frac{3t^2}{6t}=\frac{t}{2}$

For x = 1, 3t² + 1=1 ⇒ t = 0 $\Rightarrow \text{ Slope }=\frac{0}{2}=0$

Example 6: Find the equation of tangent and normal to the curve $x=\frac{2a{t}^2}{1+t^2},y=\frac{2a{t}^3}{1+t^2}$ at the point $t=\frac{1}{2}$.

Solution: Given that $x=\frac{2a{t}^2}{1+t^2}y=\frac{2a{t}^3}{1+t^2}$

at $t=\frac{1}{2},x=\frac{2a}{5},y=\frac{a}{5}$

also $\frac{dx}{dt}=\frac{4at}{{\left(1+t^2\right)}^2}\text{ and }\frac{dy}{dt}=\frac{2a{t}^2\left(3+t^2\right)}{{\left(1+t^2\right)}^2}$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{1}{2}t\left(3+t^2\right)$

when

$t=\frac{1}{2},\frac{dy}{dx}=\frac{1}{2}\frac{1}{2}\left(3+\frac{1}{4}\right)=\frac{13}{16}$

∴ The equation of the tangent when $t=\frac{1}{2}$ is 

$y-\frac{a}{5}=\left(\frac{13}{16}\right)\left(x-\frac{2a}{5}\right)$⇒ 13x – 16y = 2a

and the equation of the normal is 

$\left(y-\frac{a}{5}\right)\left(\frac{13}{16}\right)+\left(x-\frac{2a}{5}\right)=0$

⇒ 16x + 13y = 9a

Example 7: The angle of intersection between the curve, x² = 32y and y² = 4x at point (16, 8) is

(A) 60° (B) 90°

(C) ${\tan }^{-1}\left(\frac{3}{5}\right)$ (D) ${\tan }^{-1}\left(\frac{4}{3}\right)$

Ans. (C)

Solution: $x^2=32y\Rightarrow \frac{dy}{dx}=\frac{x}{16}\Rightarrow y^2=4x\Rightarrow \frac{dy}{dx}=\frac{2}{y}$

$\therefore \text{ at }(16,8),{\left(\frac{dy}{dx}\right)}_1=1,{\left(\frac{dy}{dx}\right)}_2=\frac{1}{4}$

So, required angle

$={\tan }^{-1}\left(\frac{1-\frac{1}{4}}{1+1\left(\frac{1}{4}\right)}\right)={\tan }^{-1}\left(\frac{3}{5}\right)$

9.0Tangents and Normal Practice Problems

1. The tangent to the curve y = ax² + bx at (2, –8) is parallel to x-axis. Then 

(A) a = 2, b = –2 (B) a = 2, b = –4 (C) a = 2, b = –8 (D) a = 4, b = –4

2. The slope of tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, –1) is

(A) $\frac{22}{7}$ (B) $\frac{6}{7}$ (C) –6 (D) None of these

3. If x = t² and y = 2t, then equation of the normal at t = 1 is

(A) x + y – 3 = 0 (B) x + y –1 = 0 (C) x + y + 1 = 0 (D) x + y + 3 = 0

4. The sum of intercepts on coordinate axes made by tangent to the curve is 

(A) a (B) 2a (C) $2\sqrt{a}$ (D) None of these

5. The point of the curve y² = 2(x – 3) at which the normal is parallel to the line y – 2x + 1 = 0 is:

(A) (5, 2) (B) $\left(-\frac{1}{2},-2\right)$

(C) (5, –2) (D) $\left(\frac{3}{2},2\right)$

10.0Sample Questions on Tangent and Normal

1. How do you find the slope of the tangent to a curve at a given point?

Ans: The slope of the tangent at a point (x₀, y₀) on the curve y = f(x) is the derivative f'(x₀).

2. How do you find the equation of the tangent line?

Ans: The equation of the tangent line at (x₀, y₀) is y – y₀ = m(x – x₀).

3. How do you find the equation of the normal line?

Ans: The equation of the normal line at (x₀, y₀) is y – y₀ = $-\frac{1}{m}$ (x – x₀).

4. What does it mean if the tangent is parallel to the x-axis?

Ans: The slope of the tangent is 0, i.e., f'(x₀) = 0.

5. What happens if the tangent makes equal angles with the axes?

Ans: The tangent makes angles $\theta =\frac{\pi }{4}\text{ or }\theta =\frac{3\pi }{4}$ with the x-axis, tan θ = ±1