The locus of the point of intersection of the lines, [ 2 x - y + 4 2 k = 0] and [ 2 k,x + …

Question

The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2  = 0$$ (k is any non-zero real parameter),  is :

Accepted Answer

Here, lines are :

$$\sqrt 2 x$$ - y + 4$$\sqrt 2 k$$ = 0

$$ \Rightarrow $$ $\,\,\,$$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$(i)

and$\sqrt 2 kx + ky - 4\sqrt 2  = 0\,\,\,\,\,...\left( {ii} \right) (

Put the value of y from (i) in (ii) we get;

\] \Rightarrow $2$\sqrt 2 $kx + 4$\sqrt 2 $(k2$-$1) = 0$ \Rightarrow $x =${{2\left( {1 - {k^2}} \right)} \over k}$,  y =${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$\therefore\,\,\,$${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$\therefore\,\,\,$length of transverse axis

2a = 2$ \times $4${\sqrt 2 }$= 8${\sqrt 2 }$$

Hence,  the locus is a hyperbola with  length of its transverse axis equal to 8$${\sqrt 2 }$

Solutions · 1