# Tangent - Equation of Tangent and Normal

The applications of derivatives are:

• determining the rate of change of quantities
• finding the equations of tangent and normal to a curve at a point
• finding turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs.

In this article, we will learn to use differentiation to find the equation of the tangent line and the normal line to a curve at a given point.

## Tangent and Normal Equation

We know that the equation of the straight line that passes through the point (x0, y0) with finite slope “m” is given as

y – y0 = m (x – x0)

It is noted that the slope of the tangent line to the curve f(x)=y at the point (x0, y0) is given by

$$\begin{array}{l}\frac{dy}{dx}]_{(x_{0}, y_{0})} (=f'(x_{0}))\end{array}$$

Therefore, the equation of the tangent (x0, y0) to the curve y=f(x) is

y – y0 = f ′(x0)(x – x0)

Also, we know that normal is the perpendicular to the tangent line. Hence, the slope of the normal to the curve f(x)=y at the point (x0, y0) is given by -1/f’(x0), if f’(x0) ≠ 0.

Hence, the equation of the normal to the curve y=f(x) at the point (x0, y0) is given as:

y-y0 = [-1/f’(x0)] (x-x0)

The above expression can also be written as

(y-y0) f’(x0) + (x-x0) = 0

### Points to Remember

• If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then dy/dx = slope of the tangent = tan = θ.
• If the slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x0, y0) is given by y = y0
• If θ →π/2, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at (x0, y0) is given by x = x0

### Equation of Tangent and Normal Problems

Go through the below tangent and normal problems:

Example 1:

Find the equation of a tangent to the curve y = (x-7)/[(x-2)(x-3)] at the point where it cuts the x-axis.

Solution:

As the point cut at the x-axis, then y=0. Hence, the equation of the curve, if y=0, then the value of x is 7. (i.e., x=7). Hence, the curve cuts the x-axis at (7,0)

Now, differentiate the equation of the curve with respect to x, we get

dy/dx = [(1-y)(2x-5)] / [(x-2)((x-3)]

dy/dx](7, 0) = (1-0)/[(5)(4)] = 1/20

Hence, the slope of the tangent line at (7, 0) is 1/20.

Therefore, the equation of the tangent at (7, 0) is

Y-0 = (1/20)(x-7)

20y-x+7 = 0.

Example 2:

Find the equation of tangent and normal to the curve x(⅔)+ y(⅔) = 2 at (1, 1)

Solution:

Given curve: x(⅔)+ y(⅔) = 2

Finding Equation of Tangent:

Now, differentiate the curve with respect to x, we get

(⅔)x(-⅓) + (⅔)y(-⅓) dy/dx = 0

The above equation can be written as:

dy/dx = -[y/x]

Hence, the slope of the tangent at the point (1, 1) is dy/dx](1,1) = -1

Now, substituting the slope value in the tangent equation, we get

Equation of tangent at (1, 1) is

y-1 = -1(x-1)

y+x-2 = 0

Thus, the equation of tangent to the curve at (1, 1) is y+x-2 =0

Finding Equation of Normal:

The slope of the normal at the point (1, 1) is

= -1/slope of the tangent at (1, 1)

= -1/ -1

=1

Therefore, the slope of the normal is 1.

Hence, the equation of the normal is

y-1 = 1(x-1)

y-x = 0

Therefore, the equation of the normal to the curve at (1, 1) is y-x =0

### Practice problems

Solve the following problems:

1. Calculate the slope of the tangent to the curve y=x3 -x at x=2.
2. Determine the slope of the tangent to the curve y=x3-3x+2 at the point whose x-coordinate is 3.
3. Find the equation of tangent and normal to the curve y = x3 at (1, 1).
4. Find the equation of normal at the point (am2, am3) for the curve ay2=x3.
5. The slope of the normal to the curve y=2x2 + 3 sin x at x=0 is

(a)3  (b)  -3 (c)⅓  (d) -⅓