Find the condition that one of the lines $ax^2+2hxy+by^2=0$ may coincide with one of the lines $a_1x^2+2h_1xy+b_1y^2=0$.
My Attempt:
Here, $$ax^2+2hxy+by^2=0$$ $$(\dfrac {y}{x})^{2}+\dfrac {2h}{b}.(\dfrac {y}{x})+\dfrac {a}{b}$$ Let $y=mx$ be a line represented by above equation:
Also, $$a_1x^2+2h_1xy+b_1y^2=0$$ $$(\dfrac {y}{x})^2 + \dfrac {2h_1}{b_1} (\dfrac {y}{x})+\dfrac {a_1}{b_1}=0$$ And, let $y=m_1x$ be a line represented by the above equation.
How do I move further?
Given lines are $ax^2+2hxy+by^2=0$ and $a_1x^2+2h_1xy+b_1y^2=0$
Let $y=mx$ and then we get,
First let us consider the equation $ax^2+2hxy+by^2=0$ $$ax^2+2hx(mx)+bm^2x^2=0$$ $$x^2(a+2hm+m^2b)=0$$ $$m^2b+2hm+a=0........(1)$$
Now consider the equation $a_1x^2+2h_1xy+b_1y^2=0$ $$a_1x^2+2h_1x(mx)+b_1m^2x^2=0$$ $$x^2(m^2b_1+2h_1m+a_1)=0$$ $$m^2b_1+2h_1m+a_1=0........(2)$$ Now from $(1)$ and $(2)$ we get $$\frac{m^2}{2ha_1-2ah_1}=\frac{m}{ab_1-a_1b}=\frac{1}{2bh_1-2hb_1}$$ $$m^2=\frac{ha_1-h_1a}{bh_1-hb_1}\mbox{ and }m=\frac{ab_1-a_1b}{2bh_1-2hb_1}$$ $$\left[\frac{ab_1-a_1b}{2bh_1-2hb_1}\right]^2=\frac{ha_1-h_1a}{bh_1-hb_1}$$ $$\frac{(ab_1-a_1b)^2}{4(bh_1-hb_1)^2}=\frac{ha_1-h_1a}{bh_1-hb_1}$$ $$(ab_1-a_1b)^2=4(ha_1-h_1a)(bh_1-hb_1)$$