Show that if one of the lines given by $a_1x^2+2h_1xy+b_1y^2=0$ coincides with one of the lines of $a_2x^2+2h_2xy+b_2y^2=0$ then $(a_1b_2 - a_2b_1)^2=4(a_2h_1 - a_1h_2)(b_1h_2-b_2h_1)$
Actually, I did not get any idea to start its solution. so please help me.
Help much appreciated. Thanks.
Hint: $$ ax^2+2hxy+by^2=0 \Rightarrow bm^2+2hm+a=0 $$ where $m=\frac{y}{x}$ is the slope of the straight line. So your problem is to find the condition such that the two equations:
$$ b_1m^2+2h_1m+a_1=0 \qquad b_2m^2+2h_2m+a_2=0 $$ have a common solution.
Let $\overline m$ be the common solution, and:
$$ b_1m^2+2h_1m+a_1=0 $$ has solutions $\overline m, m_1$, and $$ b_2m^2+2h_2m+a_2=0 $$ has solutions $\overline m, m_2$
We have: $$ \frac{a_1}{b_1}=\overline mm_1 \qquad \frac{a_2}{b_2}=\overline mm_2 $$ so: $$ m_1=\frac{a_1b_2}{a_2b_1}m_2 $$ and: $$ \frac{2h_1}{b_1}=-\overline{m}-m_1 \qquad \frac{2h_2}{b_2}=-\overline{m}-m_2 \quad \Rightarrow \quad \frac{2h_1}{b_1}-\frac{2h_2}{b_2}=m_2-m_1 $$ and substituting $m_1$ we find: $$ m_2= \frac{2a_2(b_2h_1-b_1h_2)}{b_2(a_2b_1-a_1b_2)} $$
since $m_2$ is a root of the second equation we must have:
$$ b_2\left( \frac{2a_2(b_2h_1-b_1h_2)}{b_2(a_2b_1-a_1b_2)}\right)^2+2h_2\left( \frac{2a_2(b_2h_1-b_1h_2)}{b_2(a_2b_1-a_1b_2)}\right)+a_2=0 $$ that, with a bit of algebra, becomes: $$ 4(b_2h_1-b_1h_2)(a_2h_1+a_1h_2)-(a_2b_1-a_1b_2)^2=0 $$