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Finding maximum number of parabola which passes through the point $A(1,2)\;,\; B(2,1)\;\;,(3,4)\;\;,(4,3)$

what i try

from these $4$ point one can imagine that parabola symmetrical about $y=x$ line

so axis of parabola along the line $y=x$ and directrix along the line $x+y+c=0$, where $c$ is any constant

did not understand how do i solve further

help me to solve it please

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    $\begingroup$ If you know nothing about the parabola, you can assume that the equation is $$ax^2+bxy+cy^2+dx+ey+f=0$$ where $b^2-4ac=0$ is one of the determining equations. Knowing four points on the parabola would be enough to find $(a,b,c,d,e,f)$ up to multiple. $\endgroup$ – Batominovski Feb 25 at 13:36
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    $\begingroup$ I just realized that you don't have a non-degenerate parabola here. The only "parabolic equations" that pass through your points are $(x-y-1)(x-y+1)=0$ and $(x+y-3)(x+y-7)=0$. Each of these two equations just give you a union of two parallel lines. $\endgroup$ – Batominovski Feb 25 at 13:55
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    $\begingroup$ The points are vertices of a rectangle. You can't inscribe a rectangle in a (nondegenerate) parabola. $\endgroup$ – Oscar Lanzi Feb 25 at 14:22
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    $\begingroup$ It can be shown that for four distinct points $A$, $B$, $C$, and $D$ on a plane, there exists a non-degenerate parabola passing through these four points if and only if they form a non-degenerate convex quadrilateral which is not a parallelogram. $\endgroup$ – Batominovski Feb 26 at 4:22
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    $\begingroup$ And, if the quadrilateral has no pair of parallel sides, there will be two nondegenerate parabolae. $\endgroup$ – Oscar Lanzi Feb 26 at 12:28
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You rightly noticed that the four points are symmetrical around $y=x$.
So change the variables accordingly $$ \left\{ \matrix{ \xi = y - x \hfill \cr \eta = y + x \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{ x = {{\eta - \xi } \over 2} \hfill \cr y = {{\eta + \xi } \over 2} \hfill \cr} \right. $$

The four points become $$ \left( {1,3} \right),\left( { - 1,3} \right),\left( {1,7} \right),\left( { - 1,7} \right) $$

The parabola's axis is the $\eta$ axis, so its formula is $$ \eta - a = k\xi ^{\,2} $$

In any case it is clear that, since the points form a rectangle, there cannot be a single parabola passing through all of them.

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Hint:

A parabola is of the form $y = ax^2 + bx + c$

We are told that it passes through the points $(1,2),(2,1)$ and $(3,4)$. (I suggest ignoring $(4,3)$ for now and revisit it later)

That means that plugging in those $x$ and $y$ values, the fact that it passes through $(\color{blue}{1},\color{red}{2})$ implies that $\color{red}{2} = a\cdot \color{blue}{1}^2 + b\cdot \color{blue}{1} + c$. Similarly we can find the equations that are implied by the parabola passing through the other points.

We have then this system of equations:

$$\begin{cases}2 = a + b + c\\1 = 4a + 2b + c\\ 4 = 9a + 3b + c\\\end{cases}$$

Now... this is a system of three linear equations and three unknowns which can by solved by standard means.

(Now, consider including $(4,3)$ as well. Does this affect the answer?)

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  • $\begingroup$ why we will take $y=ax^2+bx+c$ and why we can,take $x=ay^2+by+c$ and How can one assume that paroabola is symmetrical parallel to $x$ axis or parabola is symmetrical parallel to $y$ axis. $\endgroup$ – jacky Feb 25 at 13:39
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    $\begingroup$ Wait ... Your comment to the question noted that there's not enough information to assume symmetry about $y=x$, yet you start your answer here by asserting that "a parabola is of the form [that assumes symmetry about a vertical line]". What's up with that? $\endgroup$ – Blue Feb 25 at 13:43
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    $\begingroup$ @Blue that is a fair point. In my experience it is most common to refer to functions of $x$ in problems like these. My answer does make an assumption, one that I was not even aware of at the time, that by "parabola" we were explicitly referring to parabolic functions of $x$ which are indeed always of the form described. For conics in general who aren't necessarily functions of $x$, then as commented above one could use the more general $ax^2+bxy+cy^2+dx+ey+f=0$. $\endgroup$ – JMoravitz Feb 25 at 14:38
  • $\begingroup$ The big point though that I was trying to make was that one can convert the given information into a system of linear equations by plugging in the values of $x$ and $y$ into whatever standard form of whatever equation it is that you are interested in, be it quadratic functions of $x$, conics in general, or whatever other equation types as desired based on the specific contexts of the problem in question. $\endgroup$ – JMoravitz Feb 25 at 14:40
  • $\begingroup$ It will not be a system of linear equations. Can't be when you get two solutions, which is in fact usually the case when solutions exist. $\endgroup$ – Oscar Lanzi Feb 26 at 13:38

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