Trying to understand this solution for a high-school coordinate geometry problem better

Question

$\newcommand{\lrp}[1]{\left(#1\right)}$ The following is a high school level problem in coordinate geometry and the purpose of this post is to understand the solution better or discover other solutions or both.

Problem. Show that all chords of the curve $3x^2 - y^2 - 2x + 4y = 0$ which subtend a right angle at the origin pass through a common point.

Solution. Let $y = mx+c$ be the equation of a line not passing through the origin and which intersects the given curve at two points.
Consider the algebraic expression $$ 3x^2 - y^2 - (2x - 4y)\lrp{\frac{y-mx}{c}} $$ which is an expression of the form $Ax^2 + 2Hxy + By^2$, its zero-set is either a pair of straight lines passing through the origin (possibly not distinct) or is just the origin (which happens if $H^2 < AB$).
Further, if indeed the line $y=mx+c$ intersects the curve at two distinct points, then the zero-set of the above is necessarily the pair of of lines passing through the two points of intersection and through the origin. This is because the algebraic expression vanishes on both the points of intersection.
These lines are perpendicular to each other if $A/B = -1$, since $A/B$ is the product of the slopes of the two lines.
From here it is easy to show that all the lines $y = mx + c$ pass through a common point.

I understand this argument. But the step where we created the algebraic expression was quite a clever one and it eludes me.

Is there a "deeper reason" as to where this came from?

Also, is there any other way to prove the same? One may observe that the given curve is a hyperbola and then perhaps write down a 'hard' proof (as opposed to the 'soft' proof above). But perhaps there are other such elegant or 'more natural' proofs.

Answer 1

If $x,y$ is one intersection point, then per orthogonality $(-y/z,x/z)$ is the second point for some $z\ne 0$. This second point satisfies the equations $$ 3y^2-x^2+2yz+4xz=0\\ x = -my+cz $$ This might be confusing, so start again.

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Let the second, orthogonal intersection point be $(x_2,y_2)=(-y,x)$. Then the primary intersection point has coordinates $(x_1,y_1)=(x/z,y/z)$ for some $z\ne 0$. Then this point satisfies the equations $$ 3x^2-y^2-2xz+4yz=0\\ y=mx+cz $$ Elimination of $z$ gives the equation $$ 3x^2-y^2-2(x-2y)\frac{(y-mx)}{c}=0 $$ As this is now independent of $z$, this equation is true for all intersection points. To be valid for the claimed situation, the expanded equation $Ax^2+2Hxy+By^2=0$ has to be valid also for the second point, $Ay^2-2Hxy+Bx^2=0$, etc.

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As the coefficients are independent of the point on the ellipse, the triple $(A,H,B)$ has to be proportional to $(-B,H,-A)$, which implies $A+B=0$. Or in full form without divisors $$ 0=[3c+2m]+[-c+4]=2(m+c+2) $$ This means that the point $(x_3,y_3)=(1,-2)$ is on every such line.

Answer 2

$AB$ is one of the chord of the given curve $3x^2-y^2-2x+4y=0$, which subtends a right angle at the origin $O$. $OA$ and $OB$ are the line segments drawn from origin to the two points of intersection between the curve and the chord $AB$. We let the slope of $AB$ to be $\space m$. Since $\measuredangle AOB=90^o$, the slope of the line segment is equal to $\space -\dfrac{1}{m}$.

We start our proof by determining the coordinates of $A$ and $B$ in terms of $\space m\space$ as shown below. $$3x_A^2-m^2x_A^2-2x_A+4mx_A=0\qquad\space\rightarrow\quad x_A=\dfrac{2-4m}{3-m^2}\qquad\rightarrow\quad y_A=\dfrac{2m-4m^2}{3-m^2}$$

$$3m^2x_B^2-x_B^2-2m^2x_B-4mx_B=0\quad\rightarrow\quad x_B=\dfrac{2m^2+4m}{3m^2-1}\quad\rightarrow\quad y_B=-\dfrac{2m+4}{3m^2-1}$$

Now, we can find the equation of the chord $AB$. $$\dfrac{y-\dfrac{2m-4m^2}{3-m^2}}{x-\dfrac{2-4m}{3-m^2}}=\dfrac{y+\dfrac{2m+4}{3m^2-1}}{x-\dfrac{2m^2+4m}{3m^2-1}}$$

Simplification after cross multiplying leads to, $$\left(1+4m-m^2\right)y+\left(6+2m-6m^2\right)x=4-6m-4m^2.$$

Our aim is to show that the line representing this equation passes through a certain point for all real values of $\space m$. If this particular point is $P\space\left(x_P,y_P\right)$, then we can express its $y-$coordinate as, $$y_P=kx_P,$$ where $\space k\space$ is some real constant, because both $x_P$ and $y_P$ are also real constants. Therefore, we have, $$\left(1+4m-m^2\right)kx_P+\left(6+2m-6m^2\right)x_P=4-6m-4m^2.$$

Using this equation, It is possible to express $x_P$ in terms of the constant $\space k\space$ and the variable $\space m$. $$x_P=\dfrac{4-6m-4m^2}{6+2m-6m^2+k\left(1+4m-m^2\right)}=\dfrac{4-6m-4m^2}{\left(6+k\right)+\left(2+4k\right)m-\left(6+k\right)m^2}$$

However, we need $x_P$ to be free of $\space m$, which is possible if and only if we could find a real constant $n\space$ such that, $$n\left(4-6m-4m^2\right)= \left(6+k\right)+\left(2+4k\right)m-\left(6+k\right)m^2.$$

When we compare the coefficients of the first and the second terms of the left hand side with those of the right hand side, we get the following two equations, $$4n=6+k\qquad\text{and}\qquad 6n=-2-4k.$$

$${\large{\bf{\therefore}}}\quad n=1,\quad k=-2,\quad x_P=1, \quad\text{and}\quad y_P=-2.$$

$\underline{\text{Additional}\enspace\text{Information}}$

If a chord of a hyperbola or an ellipse given by the equation $\space ax^2+by^2+cx+dy=0\space$ subtends a right angle at the origin, it passes through the point $$\left(\dfrac{-c}{a+b},\space\dfrac{-d}{a+b}\right).$$