1
$\begingroup$

This theorem is given in the book.

Statement :

Let $$ax^2 + by^2 + 2hxy + 2fy + 2gx + c = 0$$ be a equation of a curve and let $my + nx = l, l \ne 0$ be a line intersecting the given curve at two points, then find the equation of pair of lines joining origin and the points of intersection of given curve and given line. (mind-bending statement).

Proof :

$$my + nx = l \implies {my + nx \over l} = 1$$

Therefore it easily follow from here that the desired equation is $$ax^2 + by^2 + 2hxy + 2(fy + gx)\left({my + nx \over l}\right) + c\left({my + nx \over l}\right)^2 = 0 \tag{P}$$

  • Please help me understand from where (P) come from ? Also why we are discriminating $ax^2, by^2$ and $2hxy$ by not multipling them by $\displaystyle {my + nx \over l}$ ?
  • 0
    this is simple, since the term $$\frac{my+nx}{l}=1$$ – 2017-02-20
  • 0
    We basically want a homogenous second order equation, which the points of intersection and the origin satisfy – 2017-02-20
  • 0
    @Dr.SonnhardGraubner Yes I know but what point are you trying to make ? Please elaborate. – 2017-02-20
  • 0
    @DhanviSreenivasan "We basically want a homogenous second order equation" I did know that. I did not get why this manipulation will satisfy "which the points of intersection and the origin satisfy " ? – 2017-02-20
  • 0
    you are multiplying the terms above by $1$ this is ever possible – 2017-02-20
  • 0
    @Dr.SonnhardGraubner Thanks I got the answer. – 2017-02-20

3 Answers 3

3

Please help me understand from where (P) come from ?

The author is simply expanding $1$ in the formula. You could use this as an intermediate step:

$$ ax^2+by^2+2hxy+2(fy+gx)\cdot 1 + c\cdot 1^2=0 $$

The more interesting question is that of why they are doing this. The key point is that this substitution makes all the terms quadratic. If you expand these, you get something of the form

$$\alpha x^2+\beta y^2+\gamma xy=0$$

with $\alpha,\beta,\gamma$ rational functions in $a,b,c,d,e,f,g,l,m,n$. Now the origin $x=y=0$ is definitely a solution of this equation. And the points of intersection will be solutions, too, since they satisfy both the constituent equations.

You can also see that if $(x,y)$ is a solution, then $(\lambda x,\lambda y)$ is a solution, too. So if you have a solution distinct from the origin, then the whole line joining that solution to the origin will consist of solutions, too.

Also why we are discriminating $ax^2, by^2$ and $2hxy$ by not multipling them by $\displaystyle {my + nx \over l}$ ?

That would raise the combined degree of $x$ and $y$ to more than two in parts of the formula. The result would be an algebraic curve which does contain the origin and the two points of intersection. But it would be more complicated, not simply a pair of lines. The argument about $\lambda$ I made above no longer holds in this case.

  • 0
    Thanks for the answer. But can you re-explain "So if you have a solution distinct from the origin, then the whole line joining that solution to the origin will consist of solutions, too." ? – 2017-02-20
  • 0
    @A---B: If $(x,y)$ is different from $(0,0)$, then $(\lambda x,\lambda y)$ will be another point on the line through $(0,0)$ and $(x,y)$. If you plug $(\lambda x,\lambda y)$ into the equation, you get $\lambda^2$ times the original left hand side, and thus if the left hand side was zero originally (i.e. $(x,y)$ was a solution) then $\lambda^2\cdot 0$ is zero as well, hence $(\lambda x,\lambda y)$ is a solution, too. Such an equation where the sum of degrees is the same in all terms is called homogeneous; a term used in other comments and answers. – 2017-02-20
  • 0
    Yes I got it, thanks. – 2017-02-21
1

We basically have two equations: $$ax^2 + 2hxy + by^2 + 2fy + 2gx + c = 0\space \space(1)$$ $$nx + my = l \space \space (2)$$ Now if a point satisfies (2), then, it will also satisfy $$\frac{nx+my}{l} =1 \space \space (3)$$ Now, since the point $(x,y)$ satisfies (3), it will also satisfy $$ax^2 + 2hxy + by^2 + 2(fy+gx)(\frac{nx+my}{l}) + c(\frac{nx+my}{l})^{2} = 0 $$ The above is basically a homogeneous second degree equation which contains the origin, and passes through the points of intersection of (1) and (2)

1

The first requirement:-

For $ax^2 + 2hxy + by^2 = 0$ to represent a pair of real straight lines passing through the origin we need $h^2 \ge ab$.

We assume the “a”, “b”, “h” in the equation of the conic satisfy that criteria.

When we (1) multiply only the y term by $\left({my + nx \over l}\right)$, because that factor is one, it will not increase the degree of the y term.

(2) This same is true for x term.

(3) The same is also true for the constant term (when multiplied by $(\left({my + nx \over l}\right))^2$.

After the above multiplications, (P) still satisfies the $h^2 \ge ab$ initial requirement.

The second requirement:-

For an equation to represent a pair of straight line passing through the origin, we need that equation be homogeneous in second degree in x and y.

(P), after such artificial makeup, is homogeneous.

  • 0
    So this an ad-hoc way of getting the desired results. – 2017-02-20
  • 0
    @A---B It is one way of explaining how we produce the (P). – 2017-02-20
  • 0
    Can you get the same results from a different method ? – 2017-02-20
  • 0
    @A---B Sorry no. Most textbooks explain this matter more or less the same way. They build (P) first, and then explain why they do that. – 2017-02-20