Question

The curves C1 and C2 are defined by the following parametric equations: C1: x=1+2t, y=2+3t, 2<t<5 C2: x=1/(2t-3), y=t/(2t-3), 2<t<3 Show that both curves are segments of the same straight line???

I don't understand this at all. I've work out the cartesian equation for C1 to be y=3x/2 + 1/2 and I know that to answer the question I must show the cartesian equation for C2 to be the same. However, I am not getting the same cartesian equation for C2 and cannot figure this question out at all

Answer 1

We have:

`C_1: { (x=1+2t), (y=2+3t) :} \ \ \ \ \ 2 lt t lt 5`

`C_2: { (x=1/(2t-3)), (y=t/(2t-3)) :} \ \ \ \ \ 2 lt t lt 3`

Let us derive the cartesian equations of each curve via elimination of the parameter `t`:

Consider `C_1`, by rearranging, we have:

`x=1+2t => t=(x-1)/2`

So that:

`y=2+3((x-1)/2)`
`:. 2y=4+3(x-1)`
`:. 2y=4+3x-3`
`:. 2y=1+3x`

Consider `C_2`, by rearranging, we have:

`x=1/(2t-3) => (2t-3)x=1 => t=1/2(1/x+3)`

So that:

`y=x (1/2(1/x+3))`
`:. 2y=x (1/x+3)`
`:. 2y=1+3x`

And the two cartesian equations are identical, QED

Answer 2

Kindly go through the Explanation.

Explanation:

The curve `C_1` is given by the parametric equations

`C_1 : x=1+2t, y=2+3t, t in (2,5)`.

Eliminating the parameter `t` from these eqns., we get a

relation between `x and y`, which is the cartesian eqn. of `C_1`.

We have,

`3x=3+6t and 2y=4+6t," so that, "3x-2y=-1`.

Hence, the cartesian eqn. of `C_1` is `C_1 :3x-2y+1=0`.

Likewise, for `C_2`, we have,

`3x-2y+1=3/(2t-3)-(2t)/(2t-3)+1=-1+1=0`.

Comparing the cartesian eqns. of `C_1 and C_2`, we find that

they are linear eqns. in `RR^2` of the type,

`ax+by+c=0," where, "a^2+b^2!=0`.

So, `C_1 and C_2` represent the same straight line.

BONUS :

`C_1"=the segment "AB-{A(5,8), B(11,17)}`.

`C_2"=the segment "CD-{C(1,2), D(1/3,1)}`.

`C_1 and C_2 sub" the line "{(x,y) in RR^2 : 3x-2y+1=0}`.

Enjoy Maths.!