If one of the lines in the pair of straight line given by $4 x^2+6 x y+k y^2=0$ bisects the angle between…
If one of the lines in the pair of straight line given by $4 x^2+6 x y+k y^2=0$ bisects the angle between the coordinate axes, then $k \in$
$\{-2,-10\}$
$\{-2,10\}$
$\{-10,2\}$
$\{2,10\}$
Solution
Since, one of the lines represented by $a x^2+2 h x y+k y^2=0 \quad$ bisect the angle between the axes therefore its equation is $y=x$.
Since, $y=x$ satisfies $a x^2+2 h x y+b y^2=0$, therefore
$a x^2+2 h x^2+b x^2=0$
Now for equation $4 x^2+6 x y+k y^2=0$
Here, $a=4, h=3, b=k$
Now from Eq. (i)
$\begin{array}{rlrl}
\Rightarrow & & (4+k)^2 & =4(3)^2=36 \\
\Rightarrow & & 4+k & = \pm 6 \\
\Rightarrow & k & =2,-10
\end{array}$
Hence, $k \in\{-10,2\}$
