Question

$P:2x^2-axy+6y^2=0$
$Q:3x^2-8xy+4y^2=0$

If the bisectors of the angles between the lines represented by $P$ and $Q$ are same, the value of $a$ is

• A. $8$
• B. $\frac{16}{3}$
• C. $32$
• D. $-16$

Answer 1

The correct option is C $32$

Equation of angle bisector of $2x^2-axy+6y^2=0$ is

$\frac{x^2-y^2}{2-6}=\frac{xy}{-a/2}\Rightarrow a{x}^2-a{y}^2-8xy=0$ ----------(1)

And equation of angle bisector of $3x^2-8xy+4y^2=0$ is

$\frac{x^2-y^2}{3-4}=\frac{xy}{-4}\Rightarrow 4x^2-4y^2-xy=0$ -----------(2)

From (1) and (2), we get $a=32$