First slide

Pair of straight lines

Question

The condition that one of the straight line given by the equation ${ax}^{2} + 2 hxy + {by}^{2} = 0$ may coincide with one of those given by the equation $a^{'} x^{2} + 2 h^{'} xy + b^{'} y^{2} = 0$ is

Moderate

A

${({ab}^{'} - a^{'} b)}^{2} = 4 ({ha}^{'} - h' a) (bh' - b' h)$
B

${({ab}^{'} - a^{'} b)}^{2} = ({ha}^{'} - h' a) (bh' - b' h)$
C

${(ha' - h' a)}^{2} = 4 (ab' - a' b) (bh' - b' h)$
D

${(bh' - b' h)}^{2} = 4 (ab' - a' b) (ha' - h' a)$

Solution

Let the common line be y=mx.
Then it must satisfy both the equations. Therefore, we have
${bm}^{2} + 2 hm + a = 0$ __(1)
$b^{'} m^{2} + 2 h^{'} m + a^{'} = 0$ _(2)
Solved (1) and (2), we get
$\frac{m^{2}}{2 ({ha}^{'} - h^{'} a)} = \frac{m}{{ab}^{'} - a^{'} b} = \frac{1}{2 ({bh}^{'} - b^{' h})}$
Eliminating m we get ${[\frac{{ab}^{'} - a^{'} b}{2 (2 {bh}^{'} - b^{'} h)}]}^{2} = \frac{{ha}^{'} - h^{'} a}{{bh}^{'} - b^{'} h}$
or ${({ab}^{'} - a^{'} b)}^{2} = 4 ({ha}^{'} - h^{'} a) ({bh}^{'} - b^{'} h)$

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