# Prove that the product of the length of perpendiculars from (alpha,beta) to the lines given by ax^2+2hxy+by^2=0 is (alpha^2+2halphabeta+b*beta^2)/(sqrt((a-b)^2+4h^2))?

Aug 6, 2018

Let the equations of the component straight lines are

$y + {m}_{1} x = 0 \mathmr{and} y + {m}_{2} x = 0$

So

$\left(y + {m}_{1} x\right) \left(y + {m}_{2} x\right) = {y}^{2} + \frac{2 h}{b} x y + \frac{a}{b} {x}^{2}$

$\implies {y}^{2} + \left({m}_{1} + {m}_{2}\right) x y + {m}_{2} {m}_{2} {x}^{2} = {y}^{2} + \frac{2 h}{b} x y + \frac{a}{b} {x}^{2}$

Comparing we get

${m}_{1} + {m}_{2} = \frac{2 h}{b}$

And

${m}_{1} {m}_{2} = \frac{a}{b}$

Now the product of the length of the perpendiculars from the point $\left(\alpha , \beta\right)$ is

$= \frac{\beta + {m}_{1} \alpha}{\sqrt{1 + {m}_{1}^{2}}} \cdot \frac{\beta + {m}_{2} \alpha}{\sqrt{1 + {m}_{2}^{2}}}$

$= \frac{{\beta}^{2} + \left({m}_{1} + {m}_{2}\right) \alpha \beta + {m}_{1} {m}_{2} {\alpha}^{2}}{\sqrt{1 + {m}_{1}^{2} + {m}_{2}^{2} + {m}_{1}^{2} {m}_{2}^{2}}}$

$= \frac{{\beta}^{2} + \left({m}_{1} + {m}_{2}\right) \alpha \beta + {m}_{1} {m}_{2} {\alpha}^{2}}{\sqrt{1 + {\left({m}_{1} + {m}_{2}\right)}^{2} - 2 {m}_{1} {m}_{2} + {m}_{1}^{2} {m}_{2}^{2}}}$

$= \frac{{\beta}^{2} + \frac{2 h}{b} \alpha \beta + \frac{a}{b} {\alpha}^{2}}{\sqrt{1 + {\left(\frac{2 h}{b}\right)}^{2} - 2 \frac{a}{b} + {a}^{2} / {b}^{2}}}$

$= \frac{a {\alpha}^{2} + 2 h \alpha \beta + b \cdot {\beta}^{2}}{\sqrt{{\left(a - b\right)}^{2} + 4 {h}^{2}}}$