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# Equation of Curves

A line which deviate from straightness is called curved line. Like a straight line it is also smooth and continuous. Polynomial curve with degree greater than 1 is made up of curved line; degree for any polynomial curve is highest power of variable in that polynomial. For example if there is any polynomial curve with equation 2x3 + 5x2 + 6x + 3 = 0 then degree for this curve is 3 because highest power of variable is 3. To know the equation of straight line we will need slope of line X-intercept, Y-intercept but in the case of curved line we will need the points where curve line is intersecting the X-axis and that points is also called as roots for that polynomial curve.

How to find equation of curved line in the form of ax2 + bx + c = 0

These types of line will cut the X-axis at two points by help of these two points we will be able to find the equation curved line.

Assume curved line ax2 + bx + c = 0 is intersecting the X-axis at points (α, 0) and (β, 0) then equation of curved line will be:

x2 - (α + β)x + αβ = 0

How to find equation of curved line in the form of ax3 + bx2 + cx + d = 0

These types of line will cut the X-axis at three points by help of these three points we will be able to find the equation of curved line.

Assume curved line ax3 + bx2 + cx + d = 0 is intersecting the X-axis at points (α, 0), (β, 0), (λ, 0) then equation of this curve line will be:

X3 - (α+β+λ) X2 + (αβ+βλ+λα) X - αβλ = 0

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## Parametric Form of Curve

The parametric equation is a way of define a relation using with the help of parameter. For example a curve is given by x = x(t), y = y(t) is said to be a parametrically defined curve and the functions of the curve is x = x(t) and y = y(t) are said to be the parametric equation for the curve. In the below sections, learn this concept of parametric curve equation in detail.

### Solve Parametric Curve Equations

Below are some examples on parametric curve equations

Example 1:

The line passes through the line points (4, 2) and (6, 8) when t =0 we are at the point (4, 2) and when t =1 we are at the points (6, 8) find the parametric equations.

Solution:

We write equations symbolically:

(x,y) = (1 - t) (4, 2)+ (t) (6, 8)

= (4 - 4t + 6t, 2 - 2t + 8t) = (4 + 2t, 2 + 6t)

so that

x(t) = 4 + 2t and y(t) = 2 + 6t

Example 2:

The line passes through the line points (5, 3) and (6, 7) when t =0 we are at the point (5, 3) and when t =1 we are at the points (6, 7) find the parametric equations.

Solution:

We write equations symbolically:

(x,y) = (1 - t) (5, 3)+ (t) (6, 7)

= (5 - 5t + 6t, 3 - 3t + 7t) = (5 + t, 3 + 4t)

so that the equation is

x(t) = 5 + t and y(t) = 3 + 4t

Example 3:

The line passes through the line points (6, 3) and (5, 6) when t = 0 we are at the point (6,3) and when t = 1 we are at the points (5,6) find the parametric equations.

Solution:

We write equations symbolically:

(x,y) = (1 - t) (6, 3)+ (t) (5, 6)

= (6 - 6t + 5t, 3 - 3t + 6t) = (6 + t, 3 + 3t)

so that the equation is

x(t) = 6 + t and y(t) = 3 + 3t

Example 4:

How to find the parametric equations for the line through the points (10, 8) and (15, 6) so that when t = 0 we are at the point (10, 8) and when t = 1 we are at the point (15, 6).

Solution:

The parametric equations generally written as,

To find equations

(x, t) = (1-t) (10, 8) + (t) (15, 6)

= (10-10t+15t, 8-8t+6t)

= (10+5t, 8- 2t)

(x, t)= x (t) +y (t)

(x, t) can be written as x (t) =10 +5t and y (t) =8-2t.

## Examples on Curve Equation

Below are few examples on curve line equations explained with steps:

Example 1: what will be equation of curved line which is of degree two and intersecting the X-axis at points (2, 0) and (3, 0)?

Solution: equation of curved line is of degree 2 so their equation will be

x2 - (α + β)x + αβ = 0

Here α =2 and β =3

α + β =2+3=5

αβ =2 x 3 = 6

So equation of curved line is x2 - 5x + 6 = 0

Example 2: what will be equation of curved line which is of degree three and intersecting the X-axis at points (1, 0), (2, 0) and (3, 0)?

Solution: equation of curved line is of degree 3 so their equation will be

X3 - (α+ β + λ) X2 + (αβ+βλ+λα) X - αβλ = 0

Here α =1, β =2, λ =3

α + β + λ = 1+2+3 = 6

αβ+βλ+λα = 1 x 2 + 2x3 + 3 x 1 = 2 + 6 + 3 = 11

αβλ = 1 x 2 x 3 = 6

So equation of curved line is x3 - 6x2 + 11x - 6 = 0

## Practice Problems on Equation of Curved Line

Question 1: what will be equation of curved line which is of degree two and intersecting the X-axis at points (3, 0) and (5, 0)?

Answer: x2 - 8x + 15 = 0

Question 2: what will be equation of curved line which is of degree three and intersecting the X-axis at points (1, 0), (4, 0) and (7, 0)?

Answer: x3- 12x2 + 39x + 28 = 0

### Practice Problems on Parametric Equations

Problem:

The line passes through the line points (4, 3) and (6, 4) when t =0 we are at the point (4,3) and when t =1 we are at the points (6,4) find the parametric equations.