Geometry help please parametric equation?

Question

A curve has parametric equations x=t+2, y=(1/t)-1 i) Show that the line y=(-x) does not cut the curve. ii) Show the line x+y=3 cuts the curve in just one point, state the value of the parameter at this point, and write down the coordinates of this point. iii) Show that the line 2y-x+3=0 cuts the curve in two points and find the coordinates of these points. Do not use a calculator.

Answer 1

The line y=-x written using the parameter is y=-t-2
To find the cut
-t-2 = 1/t-1 ===> t^2+t+1=0 which has no real roots.So the line doen´t cut

ii) y = 3-t-2 = 1-t So 1-t =1/t-1 so (1-t)^2/t =0 double root t=1
The point is(3,0)

iii)2y= x-3 = t-1 y= (t-1)/2

(t-1)/2 = (1-t)/t so t2 -t =2-2t t^2 +t-2 =0

t=(-1+-sqrt(1+8)) /2
= (-1+-3)/2 so t =-2 t=1

The points are (0,-3/2) and (3,0)

Answer 2

The above given solution is true and correct to the best of my knowledge and belief.

Answer 3

I would try 360/15.