If one of the straight lines given by equation $$ax^2 + 2hxy + by^2 = 0$$ coincide with one of those given by $$a_2x^2 + 2h_2xy + b_2y^2 = 0$$ and the other represented by them be perpendicular, prove that $${ha_2b_2\over b_2 - a_2} = {h_2ab\over b - a} = \sqrt{-aa_2bb_2} \tag{+}$$ All straight lines pass through origin.

Let the inclination of four lines be $m, m_2, m_3, m_4$

Now $m = m_3$ and $\displaystyle m_2 = {-1\over m_4}$

$ax^2 + 2hxy + by^2 = 0$ can be represented as $$b(y - mx)(y - m_2x) = 0$$

On expanding we get

$$b(y^2 - xy(m + m_2) + mm_2x^2) = 0$$

From which we get $-b(m + m_2) = 2h$ and $bmm_2 = a$.

Also $a_2x^2 + 2h_2xy + b_2y^2 = 0$ can be represented as $b_2(y- m_3x)(y - m_4x) = 0$

Which is same as $$b_2(y- mx)\left(y + {1\over m_2}x\right) = 0$$

On doing same procedure as we did with first pair of lines,

we get $\displaystyle 2h_2 = \left({1\over m_2} - m\right)b_2$ and $\displaystyle {-bm\over m_2} = a_2$

On substituting these values in (+) we get $${ha_2b_2\over b_2 - a_2} = {h_2ab\over b - a} = {bb_2m\over 2}$$

but $\displaystyle \sqrt{-aa_2bb_2} = b_2bm$

From this I am getting $$\sqrt{-aa_2bb_2} \ne {ha_2b_2\over b_2 - a_2} = {h_2ab\over b - a}$$.

Where did I go wrong ?

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    $\begingroup$ The proof is incomplete. Working with slopes (inclinations) doesn’t cover the case of lines parallel to the $y$-axis. $\endgroup$ – amd Feb 20 '17 at 9:24
  • $\begingroup$ @amd I think that is not necessary, since given equations for pair of lines will not include the lines parallel to y-axis $(x = a)$. If that were the case then the pair of straight equation would be $(x - a)(y - mx) = xy -mx^2 - ay + amx = 0$. No $y^2$ term. $\endgroup$ – A---B Feb 20 '17 at 9:30
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    $\begingroup$ Where does it say that $b\ne0$? We only need to have $ab-h^2\lt0$ for the equation to represent a pair of lines. By your argument, this problem doesn’t include lines parallel to the $x$-axis, either, since there’d be no $x^2$ term in that event. $\endgroup$ – amd Feb 20 '17 at 9:51
  • $\begingroup$ @amd So, what should I do to make proof complete ? Take b= 0 and solve ? $\endgroup$ – A---B Feb 20 '17 at 9:57
  • $\begingroup$ Handle the vertical case separately or find a way to prove the proposition without using the slopes of the lines. $\endgroup$ – amd Feb 20 '17 at 10:04

Where did I go wrong ?

You did nothing wrong, and $(+)$ does not hold in general.

We have

$$-b(m + m_2) = 2h\quad\text{and}\quad bmm_2 = a\quad\text{and}\quad 2h_2 = \left({1\over m_2} - m\right)b_2\quad\text{and}\quad {-b_2m\over m_2} = a_2$$

So, we get $$\frac{ha_2b_2}{b_2-a_2}=\frac{\dfrac{-b(m+m_2)}{2}\cdot \dfrac{-b_2m}{m_2}\cdot b_2}{b_2-\left(-\dfrac{b_2m}{m_2}\right)}=\frac{bb_2m}{2}$$

and $$\frac{h_2ba}{b-a}=\frac{\dfrac{(\frac{1}{m_2}-m)b_2}{2}\cdot b\cdot bmm_2}{b-bmm_2}=\frac{bb_2m}{2}$$ from which $$\frac{ha_2b_2}{b_2-a_2}=\frac{h_2ab}{b-a}$$ follows.

However, $$\frac{h_2ab}{b-a}=\sqrt{-aa_2bb_2}$$ does not hold since $$\left(\frac{h_2ab}{b-a}\right)^2=\left(\frac{b_2bm}{2}\right)^2=\frac{-aa_2bb_2}{\color{red}{4}}$$

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    $\begingroup$ If we drop the factor of 2 from the middle terms of the two original equations, the last equality does hold. $\endgroup$ – amd Feb 20 '17 at 9:22
  • $\begingroup$ @amd They are there in the question. Thanks for pointing the misprint. $\endgroup$ – A---B Feb 20 '17 at 9:26

The equations of two perpendicular lines can be put in the form $\alpha x+\beta y=0$ and $c(\beta x-\alpha y)=0$ ($c\ne0$). We now add an arbitrary third, shared line $\gamma x+\delta y=0$. The resulting equations for pairs of lines that meet the conditions of the problem are then $$\begin{align}(\alpha x+\beta y)(\gamma x+\delta y)=\alpha\gamma x^2+(\alpha\delta+\beta\gamma)xy+\beta\delta y^2&=0 \\ c(\beta x-\alpha y)(\gamma x+\delta y)=c\beta\gamma x^2+c(\beta\delta-\alpha\gamma)xy-c\alpha\delta y^2&=0\end{align}$$ from which we have $$\begin{align}a&=\alpha\gamma \\ b&=\beta\delta \\ h&= \frac12(\alpha\delta+\beta\gamma)\end{align}$$ and $$\begin{align}a_2&=c\beta\gamma \\ b_2&=-c\alpha\delta \\ h_2&=\frac12c(\beta\delta-\alpha\gamma).\end{align}$$ So, $${h_2ab\over b-a}={c(\beta\delta-\alpha\gamma)\cdot\alpha\gamma\cdot\beta\delta\over2(\beta\delta-\alpha\gamma)}=\frac12c\alpha\beta\gamma\delta$$ and $${ha_2b_2\over b_2-a_2}=-{(\alpha\delta+\beta\gamma)\cdot c\beta\gamma\cdot c\alpha\delta\over2\cdot(-c\alpha\delta-c\beta\gamma)}=\frac12c\alpha\beta\gamma\delta.$$ (The denominators are non-zero, otherwise the third line coincides with one of the two orthogonal lines.) Finally, $$aa_2bb_2 = \alpha\gamma\cdot c\beta\gamma\cdot\beta\delta\cdot (-c\alpha\delta) = -(c\alpha\beta\gamma\delta)^2$$ so $$\sqrt{-aa_2bb_2}=|c\alpha\beta\gamma\delta|.$$ Unfortunately, this doesn’t equal the previous expressions. The extra factor of 2 can be eliminated by removing it from the middle term of the original line pair equations, but without other conditions, I don’t see a way to guarantee that $c\alpha\beta\gamma\delta$ is positive.

  • $\begingroup$ +1; I agree with your last statement that we can't guarantee that $c\alpha\beta\gamma\delta > 0$ but I am following an old book which has a good tendency to skip constraints on variables. But what can you do, I can't find any modern book on analytical treatment of conic sections. If you don't mind do you have any recommendation on this topic for me ? $\endgroup$ – A---B Feb 20 '17 at 15:14

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