# Find the condition that one of the straight lines given by the equation ax^2+2hxy+by^2=0 may coincide with one of those given by the equation a_1x^2+2h_1xy+b_1y^2?

Aug 6, 2018

Given equations of pair of straight lines are

$a {x}^{2} + 2 h x y + b {y}^{2} = 0. \ldots . . \left(1\right)$

and

${a}_{1} {x}^{2} + 2 {h}_{1} x y + {b}_{1} {y}^{2} = 0. \ldots . . \left(2\right)$
Obviously the straight lines are passing through orin as suggested by the equations.

Now let $\textcolor{red}{y = m x \ldots . \left(3\right)}$ be the equation of the line which coincides.

So by (1) and (3) we get

$a {x}^{2} + 2 h x \cdot m x + b {m}^{2} {x}^{2} = 0$

$\textcolor{b l u e}{\implies b {m}^{2} + 2 h m + a = 0. \ldots . . \left(4\right)}$

Similarly by (2) and (3) we will get

${a}_{1} {x}^{2} + 2 {h}_{1} x \cdot m x + {b}_{1} {m}^{2} {x}^{2} = 0$

$\textcolor{m a \ge n t a}{\implies {b}_{1} {m}^{2} + 2 {h}_{1} m + {a}_{1} = 0. \ldots . . \left(5\right)}$

By cross multiplication of (4) and (5) we get

${m}^{2} / \left(2 h {a}_{1} - 2 {h}_{1} a\right) = \frac{m}{a {b}_{1} - {a}_{1} b} = \frac{1}{2 {h}_{1} b - 2 h {b}_{1}}$

Hence we get two values of $m$

(1) $m = \frac{2 \left(h {a}_{1} - {h}_{1} a\right)}{a {b}_{1} - {a}_{1} b}$
and
(2) $m = \frac{a {b}_{1} - {a}_{1} b}{2 \left({h}_{1} b - h {b}_{1}\right)}$

By the condition of the problem these two values of $m$ must be same.

Hence

(2(ha_1-h_1a))/(ab_1-a_1b)= (ab_1-a_1b)/(2(h_1b-hb_1))

$\textcolor{m a \ge n t a}{\implies 4 \left(h {a}_{1} - {h}_{1} a\right) \left({h}_{1} b - h {b}_{1}\right) = {\left(a {b}_{1} - {a}_{1} b\right)}^{2}}$

This is the required condition.