Suppose the pairs of straight lines $x^2-2 a x y-y^2=0$ and $x^2-2 b x y-y^2=0$ are such that each pair…
Suppose the pairs of straight lines $x^2-2 a x y-y^2=0$ and $x^2-2 b x y-y^2=0$ are such that each pair bisects the angles between the other two. Then $a b=$
$1$
$-1$
$2$
$\frac{1}{2}$
Solution
Given pairs of straight lines $x^{2-2 a x y}$ and $x^2$ $2 b x y-y^2=0$ bisect the angles between the other two.
For $x^2-2 a x y-y^2=0$ equation of pairs of +1 straight lines which istobe bisected will be
$x^2+2 x y-a y^2=0$
on complaining it with
$x^2-2 a b x y-y^2=0$
We get $a=1, b=1$
now $a, b=-1$