Question

Show that the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents two parallel lines, if $b{g}^2=a{f}^2$ and $h^2=ab$ and distance betwen these parallel lines is $2\sqrt{\frac{g^2-ac}{a(a+b)}}$.

Answer 1

Given $a{x}^2+2hxy+b{y}^2+2yx+2by+c=0$

let the two parallel line is given by

$L_1:a_{1}x+b_{1}y+c_1=0$

$L_2:a_{1}x+b_{1}y+c_2=0$

$(a_{1}x+b_{1}y+c_1)(a_{1}x+b_{1}y+c_2)=a{x}^2+2hxy+b{y}^2+2gx+2by+c$

$\Rightarrow a_1^2{x}^2+2\left(a_1{b}_1\right)xy+b_1^2{y}^2+a_1(c_1+c_2)x+b_1(c_2+c_1)y+c_1{c}_2=a{x}^2+2hxy+b{y}^2+2gx+2by+c$

coefficient of $x^2:a_1^2=a$ ...(i)

coefficient of $y^2:b_1^2=b$ ...(ii)

coefficient of $xy:a_1{b}_1=h$

coefficient of $x:a_1(c_1+c_2)=2g$

coefficient of $y:b_1(c_2+c_1)=2f$

constant term : $c_1{c}_2=c$

multiplying (i) & (ii)

$ab=a_1^2{b}_1^2=h^2$

$\therefore h^2=ab$ (proved)

$a{b}^2=a_1^2\frac{b_1^2}{4}(c_1+c_2{)}^2=b{y}^2$

$\therefore a{f}^2=b{g}^2$ (proved)

$d=\frac{|c_2-c_1|}{\sqrt{a_1^2+b_1^2}}=\frac{\sqrt{\frac{4g^2}{a_1^2}-4c}}{\sqrt{a+b}}$

$\therefore d=2\sqrt{\frac{g^2-ac}{a(a+b)}}$ (proved)