# The equation of a curve is y^2+2x = 13 and the equation of a line is 2y+x = k, where k is a constant.(i) In the case where k = 8, find the coordinates of the points of intersection of the line and...

The equation of a curve is y^2+2x = 13 and the equation of a line is 2y+x = k, where k is a constant.

(i) In the case where k = 8, find the coordinates of the points of intersection of the line and the curve.

### 2 Answers | Add Yours

`y^2 + 2x = 13 ------(1) rarr y^2 = 13-2x -----(1)'`

`2y+x = k--------(2)`

When k=8

`2y+x = 8`

`2y = 8-x`

`y^2 = (8-x)^2/4` ---------------(3)

At points when curve intersects line `(1)' = (3)`

`13-2x = (8-x)^2/4`

`52-8x = x^2 -16x +64`

`0=x^2-8x+12`

By solving the above we will get x = 6 and x = 2

When x= 6 then y = 1

When x= 2 then y = 3

*So the intersection points are (6,1) and (2,3)*

The intersection between a line and a curve can be found by solving the system of equations of line and curve.

Put k = 8 and determine the equation of line:

`2y+x = 8`

Take out x from this equation:

`x = 8 - 2y`

Put 8 - 2y in equation of curve and you will have an equation in one variable:

`y^2 + 2(8 - 2y) = 13`

`y^2 + 16 - 4y - 13 = 0`

`y^2 - 4y + 3 = 0 => y^2 - (3+1)y + 3 = 0`

`y^2 - 3y - y + 3 = 0 => (y^2-3y) - (y-3) = 0`

Factorization is required:

`y(y-3)-(y-3) = 0`

`(y-3)(y-1) = 0`

This product is zero if one of members is zero:

`y-3 = 0`

or

`y-1 = 0`

Put constants to the right:

`y =3`

or

`y =1`

Find x:

`x = 8 - 2*3 => x = 2`

`x = 8 - 2 => x = 6`

**Solutions of system are points of intersection between curve and line: (2,3) and (6,1).**