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Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.


Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows; here. $$ax^2+2hxy+by^2=0$$ Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides : $$ax+hy=\pm y\sqrt{h^2-ab}.$$

What should I do next?

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If $ax^2+2hxy+by^2=0$ bisects the coordinate axes then the points $(x,x)$ and $(x,-x)$ belong to these lines. If we consider $(x,x)$ we get $$ax^2+2hx^2+bx^2=0$$ and if we consider $(x,-x)$ we get $$ax^2-2hx^2+bx^2=0.$$ If $x\ne 0$ then it is $$a+b=\pm 2h\implies (a+b)^2=4h^2.$$

Conversely, if $4h^2=(a+b)^2$ then, if $2h=a+b$ we have $ax^2+2hxy+by^2=(x+y)(ax+by)$ and if $2h=-(a+b)$ we have $ax^2+2hxy+by^2=(x-y)(ax+by).$ In any case, $ax^2+2hxy+by^2=0$ contains one of the lines $x+y=0$ or $x-y=0,$ which bisect the coordinate axis.

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  • $\begingroup$ how did you get $$a+b=\pm2h$$ $\endgroup$ – Iaamuser user Feb 9 '16 at 17:32
  • $\begingroup$ If $(a+b)^2=4h^2$ then $|a+b|=\sqrt{(a+b)^2}=\sqrt{4h^2}=2|h|,$ from where it follows. $\endgroup$ – mfl Feb 9 '16 at 17:34
  • $\begingroup$ how did you write.directly $a+b=\pm2h$ after you consider $x\ne 0$? $\endgroup$ – Iaamuser user Feb 9 '16 at 17:39
  • $\begingroup$ If $x\ne 0$ in $ax^2+2hx^2+bx^2=0$ we can divide both terms of the equality by $x^2$ to get $a+2h+b=0.$ That is, $a+b=-2h.$ $\endgroup$ – mfl Feb 9 '16 at 17:47
  • $\begingroup$ ok, i got that! $\endgroup$ – Iaamuser user Feb 9 '16 at 17:48
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$$(a+b)^2=4h^2 \Rightarrow a+b=\pm 2h$$ Let $$ax^2+2hxy+by^2=0$$ be factorized as $$(y-m_1x)(y-m_2x)=0$$ Then, $$m_1+m_2=-\frac{2h}{a}$$ $$m_1m_2=\frac{b}{a}$$ We have, $$1+\frac{b}{a}=\pm \frac{2h}{a}$$ So, $$1+m_1m_2=\pm (m_1+m_2)$$ If either of $m_1$ or $m_2$ is zero, the other must be $\pm 1$, which means we are done.

If none of them are zero, let $m_1=km_2$. $$1+km_2^2=\pm m_2(1+k)$$ $$km_2^2 \pm m_2(1+k) +1=0$$ which gives $m_2=\pm 1, \pm \frac{1}{k}$ which means $m_1=\pm \frac{1}{k},\pm 1$ correspondingly.

Thus at least one of them must bisect the coordinate axes.

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I think this one is easier to work from the other direction. If $(a+b)^2 = 4 h^2$, then $2 h = \pm (a + b)$. Thus, the equation for the lines becomes $$ a x^2 \pm (a + b) xy + b y^2 = 0 \quad \Rightarrow \quad (ax \pm by)(x \pm y) = 0. $$ Since $(x \pm y)$ is one of the factors, the line $y = \pm x$ will be a solution to the equation.

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Let $y=mx$ be the required line. Then we can rewrite the given pair of lines as $$bm^2+2hm+a=0$$.

$$\implies m=\frac{-2h\pm\sqrt{4h^2-4ab}}{2b}$$

$$(a+b)^2=4h^2 \implies 4h^2-4ab=(a-b)^2$$

Substituting in the value of $m$, $$m=\frac{-\vert a+b\vert\pm\vert a-b\vert}{2b}$$

For all values of $a,b$, the above will always gives at least $1$ or $-1$.

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Well... It seems it is asked to demonstrate that it works if $(a+b)^2=4h^2$, not if and only if...

So if you replace $h$ with $\pm \dfrac{a+b}{2}$ in your first equation you get the following:

$(x\pm y)(ax\pm by)=0$, which is your answer...

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If one of the line of $$ax^2 +2hxy+by^2=0 \tag{1}$$ bisects the coordinate axes then the eqn. of that line might be $y=x$ OR, $y =-x$. Combining these eqns we get $y=\pm x$ and substituting this value of $y$ in eqn. (1) we get \begin{eqnarray} ax^2+2hx(\pm x)+by^2=0 \\ ax^2±2hx^2+by^2=0 \\ x^2(a±2h+b)=0 \\ a±2h+b=0 \\ a+b=±2h \end{eqnarray}

Also, $(a+b)^2=4h^2$

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