When are the points of intersection of two real affine conics orthogonal to each other?

Question

Show that the lines joining the origin to the points of intersection of two curves $ax^2+2hxy+by^2+2gx=0$ and $a_1x^2+2h_1xy+b_1y^2+2g_1x=0$ will be at right angles to one another if $g(a_1+b_1)=g_1(a+b)$.

My approach:

$$\begin{eqnarray} \tag {1} ax^2+2hxy+by^2+2gx=0 \\ \tag {2} a_1x^2+2h_1xy+b_1y^2+2g_1x=0 \end{eqnarray}$$

From equation (2),

$$2x=\frac {a_1x^2+2h_1xy+b_1y^2}{g_1}$$

I could not move on from here. Please help me to continue.

A good first step looks to be eliminating $xy$ from one of the equations. I.e use the second equation to replace $xy$ in the first equation. — Zestylemonzi, Sep 10 '16 at 08:31
@Zestylemonzi, After doing so, I got $$ax^2-\frac {a_1hx^2+b_1hy^2+2g_1hx}{h_1} + by^2+2gx=0$$,
what should I do next? — pi-π, Sep 10 '16 at 10:21

David Quinn · Accepted Answer · 2016-09-10T11:05:15.433

2

You need first to eliminate the $x$ term from the equations, so that the equation is homogeneous. Multiply the first equation by $g_1$ and the second equation by $g$ and subtract.

You then have $$x^2(g_1a-ga_1)+2xy(g_1h-gh_1)+y^2(g_1b-gb_1)=0$$

The gradient from the origin to $(x,y)$ is $\frac yx=m$ so we divide by $x^2$ and write this equation as a quadratic in $m$.

For the perpendicularity condition we require the product of the roots to be $-1$, so putting that in terms of the coefficients, we get $$\frac{g_1a-ga_1}{g_1b-gb_1}=-1$$

This leads immediately to the result you are looking for.

I hope this helps

edited Sep 10 '16 at 11:05

answered Sep 10 '16 at 10:55

David Quinn

34,121

Do we need to subtract second from first after multiplication? – pi-π Sep 10 '16 at 11:09
Can we directly replace $\frac {y}{x}$ with $m$ without giving any reason? – pi-π Sep 10 '16 at 11:15
Yes you need to get rid of the $x$ term. and yes it should be clear why we are using $m$. – David Quinn Sep 10 '16 at 11:23
Could.you.please give me the exact reason so that I.could use $m$ in place of $\frac {y}{x}$? – pi-π Sep 10 '16 at 11:26
Its just a substitution if you like! But it represents the gradient of the line. Use another letter if you prefer. – David Quinn Sep 10 '16 at 11:31
If I use another letter then again I.had to use it for the perpendicularity condition.
Moreover, I could not understand the perpendicularity condition. How did you get that?
– pi-π Sep 10 '16 at 11:33
$m_1\bot m_2\iff m_1m_2=-1$ – David Quinn Sep 10 '16 at 12:47

When are the points of intersection of two real affine conics orthogonal to each other?

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