Question

$P:x^2-y^2+2y-1=0$
$L:x+y=3$

Combined equation of L and L' is:

• A. $x^2-y^2-4x+2y+3=0$
• B. $x^2-y^2-2x+4y-3=0$
• C. $x^2-y^2+2x-4y-3=0$
• D. $x^2-y^2+4x-2y+3=0$

Answer 1

The correct option is B $x^2-y^2-2x+4y-3=0$

$x^2-y^2+2y-1=0$
$x^2-(y^2-2y+1)=0$
$x^2-(y-1{)}^2=0$
$(x-(y-1\left)\right)(x+y-1)=0$
$(x-y+1)(x+y-1)=0$
Thus the lines are
$x-y+1=0$
$x+y-1=0$
Adding both, we get
$2x=0$
$x=0$
Substitution in any one equation gives us
$y=1$.
Thus the point of intersection is $(0,1)$.
Now
$L:x+y=3$
Since $L^{\prime }$ is perpendicular to L
Therefore
$L^{\prime }:x-y=d$
It passes through $(0,1)$
$\therefore d=-1$
Hence, $L^{\prime }:x-y+1=0$
Thus the combined equation of L and L' is
$(x-y+1)(x+y-3)=0$
$x^2-y^2-3x+3y+x+y-3=0$
$x^2-y^2-2x+4y-3=0$