# Find loci (parameter involved)

• September 16th 2011, 11:04 PM
piscoau
Find loci (parameter involved)
Two straight line: kx+y=1-2k, x-k(y-2)-1. Let P be the point of intersection of these 2 lines. Find the equation of locus of P when the value of k varies.

I found that k = (x-y)/(x+y), x= -(k+1)(2k-1)/(k^2+1), y= (k-1)(2k-1)/(k^2+1).
But I can't find a good method to eliminate the parameter k in above two equations.
Could any one help me?
• September 17th 2011, 04:35 AM
FernandoRevilla
Re: Find loci (parameter involved)
Quote:

Originally Posted by piscoau
Two straight line: kx+y=1-2k, x-k(y-2)-1. Let P be the point of intersection of these 2 lines. Find the equation of locus of P when the value of k varies.

Supposing the lines are $kx+y=1-2k$ and $x-k(y-2)-1=0$ then, $k=\frac{1-y}{x+2}=\frac{1-x}{y-2}\Leftrightarrow \ldots \Leftrightarrow x^2-y^2+x+3y-4=0$ (hyperbola).