Question

Find the equation of straight line having a gradient of -`1/t` and passing through the point(`t^2`,2t).(pls see details below). ?

This line meets the line 2y+x=4+2`t^2` at the point P.
Show that the x-coordinates of P is 2t(t-1) and find the y-coordinates of P.
Find the locus of P as t varies.

Answer 1

See below

Explanation:

If you know the slope `m` (also known as gradient) of a line, and one its points `(x_0,y_0)`, the equation of the line is given by

`y-y_0=m(x-x_0)`

In your case, `m=-1/t`, and `(x_0,y_0)=(t^2,2t)`. So, the equation is

`y-2t = -1/t(x-t^2)`

we may rewrite this equation as

`y = -1/tx+3t`

The other line is `2y+x=4+2t^2`, which we can write as

`y = 2+t^2-x/2`

(I just solved for `y` bringing everything else to the right and dividing by `2`).

To find the point where the lines meet, let's ask for both equation to be satisfied. Since we know the expression for `y` for both lines, let's set them to be equal:

`-1/tx+3t = 2+t^2-x/2`

Solving for `x`, we get:

`-1/tx+x/2 = t^2-3t+2`

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

`x(\frac{t-2}{2t})= t^2-3t+2`

`x= \frac{2t}{t-2}(t^2-3t+2)`

The last thing we need to do is to observe that `t^2-3t+2` can be factored as

`t^2-3t+2 = (t-1)(t-2)`

and thus the expression becomes

`x = \frac{2t}{cancel(t-2)}(t-1)cancel((t-2))=2t(t-1)`

as required.