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.

1 Answer
KillerBunny
Jun 25, 2018

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.

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