Question

If one of the straight lines given by the equation $a{x}^2+2hxy+b{y}^2=0$ coincide with one of those given by $a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2=0$ and the other lines represented by them be perpendicular,

then prove that $\frac{h{a}^{\prime }{b}^{\prime }}{b^{\prime }-a^{\prime }}=\frac{h^{\prime }ab}{b-a}=\frac{1}{2}\sqrt{-a{a}^{\prime }b{b}^{\prime }}$

Answer 1

Let the lines represented by $a{x}^2+2hxy+b{y}^2=0$ be $y=m_{1}x$ and $y=m_{2}x$

Then $m_1,m_2$ are the roots of the equation $b{m}^2+2hm+a=0$

$m_1+m_2=\frac{-2h}{b}.......\left(i\right)m_1{m}_2=\frac{a}{b}........\left(ii\right)$

According to the problem the lines represented by $a{x^{\prime }}^2+2h{x}^{\prime }{y}^{\prime }+b{y^{\prime }}^2=0$

are $y=m_{1}x$ and $y=-\frac{1}{m_2}x$

$m_1+(-\frac{1}{m_2})=\frac{-2h^{\prime }}{b}....\left(iii\right)$

$m_1.(-\frac{1}{m_2})=\frac{a^{\prime }}{b^{\prime }}....\left(iv\right)$

Multiplying $\left(ii\right)$ and $\left(iv\right)$

${\Rightarrow m}_1^2=-\frac{a{a}^{\prime }}{b{b}^{\prime }}{m}_1=\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{b{b}^{\prime }}$

Substituting in $\left(ii\right)$

$\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{b{b}^{\prime }}.m_2=\frac{a}{b}\Rightarrow m_2=\frac{a{b}^{\prime }}{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}=\frac{a{b}^{\prime }\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{-a{a}^{\prime }b{b}^{\prime }}=-\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{a^{\prime }b}$

Substituting values of $m_1$ and $m_2$ in $\left(i\right)$

$\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{b{b}^{\prime }}-\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{a^{\prime }b}=\frac{-2h}{b}\sqrt{-a{a}^{\prime }b{b}^{\prime }}(\frac{1}{b^{\prime }}-\frac{1}{a^{\prime }})=-2h\frac{1}{2}\sqrt{-a{a}^{\prime }b{b}^{\prime }}=\frac{h{a}^{\prime }{b}^{\prime }}{a^{\prime }-b^{\prime }}......\left(v\right)$

Substituting values of $m_1$ and $m_2$ in $\left(iii\right)$

$\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{b{b}^{\prime }}-\frac{\sqrt{-a{a}^{\prime }b{b}^{\prime }}}{a{b}^{\prime }}=\frac{-2h^{\prime }}{b^{\prime }}\frac{1}{2}\sqrt{-a{a}^{\prime }b{b}^{\prime }}(\frac{1}{b}-\frac{1}{a})=-2h^{\prime }\Rightarrow \frac{h{a}^{\prime }{b}^{\prime }}{a^{\prime }-b^{\prime }}=\frac{h^{\prime }ab}{a-b}........\left(vi\right)$

From $\left(v\right)$ and $\left(vi\right)$

$\frac{h^{\prime }ab}{a-b}=\frac{h{a}^{\prime }{b}^{\prime }}{a^{\prime }-b^{\prime }}=\frac{1}{2}\sqrt{-a{a}^{\prime }b{b}^{\prime }}$

Hence proved.