Prove that two of the lines represented by the equation $$ay^4+bxy^3+cx^2y^2+dx^3y+ex^4=0$$will be perpendicular if $$(b+d)(ad+be)+(e-a)^2(a+c+e)=0$$

I tried to solve the equation by assuming two arbitrary pairs of line $$(ay^2+ex^2+2hxy)(x^2-pxy+y^2)$$ so as to make the second pair that of perpendicular lines. (I assumed $h$ and $p$ arbitrarily)

I then multiplied the terms in the bracket and by comparing the terms found the value of $h$ and $p$, but I couldn't get any equation which would lead to the required condition.

Can anybody just give me a small hint as to how can i advance further.

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    $\begingroup$ Probable typo in the last term. $\endgroup$ – Yves Daoust Aug 31 '16 at 6:43
  • $\begingroup$ Or in the first? $\endgroup$ – lab bhattacharjee Aug 31 '16 at 6:43
  • $\begingroup$ @labbhattacharjee: quite right. $\endgroup$ – Yves Daoust Aug 31 '16 at 6:44
  • $\begingroup$ I just checked by question, it is mentioned here just the way its in my book. $\endgroup$ – Harsh Sharma Aug 31 '16 at 6:44
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    $\begingroup$ We did, didn't we ? $\endgroup$ – Yves Daoust Aug 31 '16 at 6:46

1) First of all, your factorization into two homogeneous degree 2 polynomials, the second one taking into account for the equations of the straight lines is a good idea, and I start from it, but it should be:



  • in the second factor, a minus sign in front of $y^2$ instead of a plus sign.

(as results from expansion of $(x-\alpha y)(x+\frac{1}{\alpha} y)$).

  • in the first factor, $-ay^2$ instead of $ay^2$ in order to get $ay^4$ in the expansion.

2) Expanding (0) and identifying coefficients of $x^2y^2$, $x^3y$ and $xy^3$, resp., gives the system:

$$\begin{cases}a+e+2hp&=&-c&(1) \\2h-ep&=&d&(2)\\-2h+ap&=&b& (3)\end{cases}$$

It is now a matter of elimination of $p,q$ between these 3 equations.

More precisely, (2) and (3) give explicit expressions of $h$ and $p$ as functions of $a,b,d,e$:

$$h=\dfrac{ad + be}{2(a-e)} \ \ \text{and} \ \ p=\dfrac{b+d}{a-e}$$

When these expressions are plugged in (1), one obtains the looked for identity.

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    $\begingroup$ I had made an error that I have corrected. I wanted to start from your own factorization that I had assumed exact. In fact 2 changes had to be made. $\endgroup$ – Jean Marie Aug 31 '16 at 8:45


The slopes $y/x$ of the two perpendicular lines can be written $m$ and $-1/m$. When you plug this in the given equation, you get the conditions

$$\begin{cases}am^4+bm^3+cm^2+dm+e=0,\\ a-bm+cm^2-dm^3+em^4=0.\end{cases}$$

Now you must eliminate $m$ from these. Try to compute the $\gcd$ of the two polynomials by the Euclidean algorithm.



As $y=0\iff x=0$

let us divide the equation by $x^4$ and write $m=\dfrac yx$ to find $$am^4+bm^3+cm^2+dm+e=0$$

Clearly $m$ represents the gradient of the straight lines.

Using this, $$am^4+bm^3+cm^2+dm+e=0=a(m^2+pm-1)(m^2+qm-1)$$

Now we need to eliminate $m$ by equating the constants and the coefficients of the different powers of $m$


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