The bisector of two line L1 and L2 are given by 3x^2-8xy-3y^2+10x+20y-25 If the line L1 passes through the origin - Maths - Conic Sections

Question

The bisector of two line L1 and L2 are given by
3x^2-8xy-3y^2+10x+20y-25 If the line L1 passes through the origin
find L2???

Himanshu · Accepted Answer

Let line L1 be y = mx

Let line L2 be ax + by + c = 0

Equation of bisector of angle made by the L1 and L2 will be given by

$\frac{y - m x}{\sqrt{1 + m^{2}}} = \pm \frac{a x + b y + c}{\sqrt{a^{2} + b^{2}}} \Rightarrow \frac{(y - m x)^{2}}{1 + m^{2}} = \frac{(a x + b y + c)^{2}}{(a^{2} + b^{2})} = \frac{a^{2} x^{2} + b^{2} y^{2} + c^{2} + 2 a b x y + 2 a x c + 2 b y c}{a^{2} + b^{2}} \Rightarrow (\frac{1}{1 + m^{2}} - \frac{b^{2}}{a^{2} + b^{2}}) y^{2} + (\frac{m^{2}}{1 + m^{2}} - \frac{a^{2}}{a^{2} + b^{2}}) x^{2} - x y (\frac{2 m}{1 + m^{2}} + \frac{2 a b}{a^{2} + b^{2}}) - \frac{2 a c x}{a^{2} + b^{2}} - \frac{2 b c y}{a^{2} + b^{2}} - \frac{c^{2}}{a^{2} + b^{2}} = 0$

$\frac{2 a c}{a^{2} + b^{2}} = - 10 a n d \frac{2 b c}{a^{2} + b^{2}} = - 20 \Rightarrow 4 c^{2} = (- 10)^{2} + (- 20)^{2} = 500 \Rightarrow c = \pm 5 \sqrt{5} \Rightarrow \frac{a}{b} = \frac{1}{2} \frac{c^{2}}{a^{2} + b^{2}} = 25 \Rightarrow a^{2} + b^{2} = 5 \Rightarrow b = \pm 2 a n d a = \pm 1$

$x + 2 y + 5 \sqrt{5} = 0$

The bisector of two line L1 and L2 are given by 3x^2-8xy-3y^2+10x+20y-25.If the line L1 passes through the origin find L2???