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.

**Also, read:**

## Tangent and Normal Equation

We know that the equation of the straight line that passes through the point (x_{0}, y_{0}) with finite slope “m” is given as

y – y_{0} = m (x – x_{0})

It is noted that the slope of the tangent line to the curve f(x)=y at the point (x_{0}, y_{0}) is given by

Therefore, the equation of the tangent (x_{0}, y_{0}) to the curve y=f(x) is

**y – y**_{0}** = f ′(x**_{0}**)(x – x**_{0}**)**

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 (x_{0}, y_{0}) is given by -1/f’(x_{0}), if f’(x_{0}) ≠ 0.

Hence, the equation of the normal to the curve y=f(x) at the point (x_{0}, y_{0}) is given as:

y-y_{0} = [-1/f’(x_{0})] (x-x_{0})

The above expression can also be written as

**(y-y**_{0}**) f’(x**_{0}**) + (x-x**_{0}**) = 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 (x
_{0}, y_{0}) is given by y = y_{0} - 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 (x
_{0}, y_{0}) is given by x = x_{0}

### 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]^{⅓}

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:

- Calculate the slope of the tangent to the curve y=x
^{3}-x at x=2. - Determine the slope of the tangent to the curve y=x
^{3}-3x+2 at the point whose x-coordinate is 3. - Find the equation of tangent and normal to the curve y = x
^{3}at (1, 1). - Find the equation of normal at the point (am
^{2}, am^{3}) for the curve ay^{2}=x^{3}. - The slope of the normal to the curve y=2x
^{2}+ 3 sin x at x=0 is

(a)3 (b) -3 (c)⅓ (d) -⅓

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