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Find the equations of a pair of straight lines which pass through the origin and are perpendicular to each of the lines represented by $ax^2+2hxy+by^2=0$.

I have already posted this question few days before but could not get answer. My book shows $bx^2-2hxy+ay^2=0$ as answer, but I didn't get that answer.

Some Guy
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Ger Wyn
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    You are under the misapprehension that this site is a problem help-line. It is far more likely that you will get assistance here if you show what you tried to do that "didn't work". Questions that only give the statement of a problem with no discussion of what you attempted to do are generally going to get down-voted and/or closed. – colormegone Jan 04 '16 at 17:33
  • @RecklessReckoner Fully agree, and yet people give full solutions ! – Shailesh Jan 07 '16 at 05:32

1 Answers1

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Let $y=m_1x$ & $y=m_2 x$ be the lines represented by $ax^2+2hxy+by^2=0$ or $\frac{a}{b}x^2+\frac{2h}{b}xy+y^2=0$ then one should have $$(y-m_1x)(y-m_2x)=\frac{a}{b}x^2+\frac{2h}{b}xy+y^2 $$ $$m_1m_2 x^2-(m_1+m_2)xy+y^2=\frac{a}{b}x^2+\frac{2h}{b}xy+y^2$$ comparing the corresponding coefficient, one should get $$m_1m_2=\frac ab\tag 1$$ $$m_1+m_2=-\frac{2h}b\tag 2$$ Now, let $y=-\frac{x}{m_1}$ & $y=-\frac{x}{m_2}$ be the pair of lines perpendicular to the pair of lines: $ax^2+2hxy+by^2=0$ & passing through the origin then the combined equation is given as $$\left(y+\frac{x}{m_1}\right)\left(y+\frac{x}{m_2}\right)=0$$ $$x^2+(m_1+m_2)xy+m_1m_2y^2=0$$ setting the values of $m_1m_2$ & $(m_1+m_2)$ from (1) & (2), one should get $$x^2+\left(-\frac{2h}{b}\right)xy+\frac aby^2=0$$ $$\color{red}{bx^2-2hxy+ay^2=0}$$

Harish Chandra Rajpoot
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