# How reconstitute a conic equation by 4 roots?

1. Sep 29, 2015

### Bruno Tolentino

Given two points in x-axis, a and b, is possible formulate a quadratic equation whose roots intersects a and b: (x-a)(x-b)=0

So, is possible make the same with a conic equation? Given 4 points, two in x-axis (a and b) and two in y-axis (c and d), is possible reconstitute a conic equation whose roots in x and y are a, b, c and d?

2. Sep 29, 2015

### HallsofIvy

Not uniquely. This is $x^2- (a+ b)x+ ab= 0$. Multiplying the entire equation by any non-zero number gives another quadratic equation having the same roots.
If you want the quadratic function with leading coefficient 1, given by $y= x^2+ bx+ c$ then you need to determine two values, a and b. That requires two linear equations which you can get by replacing x and y with values of two points, not just where the graph crosses the x-axis.

A general conic is of the form $ax^2+ bxy+ cy^2+ dx+ ey+ f= 0$. Even assuming a= 1 we have five coefficients, b, c, d, e, and f, to determine. That will require five points, not just four.

3. Sep 29, 2015

### Bruno Tolentino

I want ANY conic equantion that intersects a, b, c and d! And, more importante: I want to write this conic in the format (x-a)(x-b) + (y-c)(y-d) + (x-a)(x-d) = 0! Or, in some format like this...

4. Sep 29, 2015

### HallsofIvy

Since you know what "conics" are, surely you know that the graphs of conics are one of:
ellipse,
circle,
parabola,
hyperbola,
a single straight line,
two parallel straight lines,
two intersecting straight lines.
(The last three are "degenerate conics")

NONE of those cross the x-axis (or any horizontal straight line) more than twice so what you are asking for is impossible.

5. Sep 29, 2015

### Staff: Mentor

I don't believe HallsOfIvy read your question correctly. Given two x-intercepts a and b, and two y-intercepts c and d, there are many conics that meet this requirement: circles, ellipses, and hyperbolas. The only conic that doesn't satisfy the requirements is a parabola.