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This is the work I've done so far. I plugged in the x and y values into each conic's standard form equation. But I'm not sure where to go from here. I've been really confused with conics.

Ellie
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  • Our teacher gave us the hint: rotate the given points 45 degrees counterclockwise around the origin, then determine the equation of the parabola – Ellie Nov 01 '21 at 16:43
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    Do you know that given any three points in a plane you can either draw a unique straight line or a unique circle through them? How many parameters do you have in defining your conic? How many points do you have to pass through? What might the general case be, and what conditions might cause a problem? – Mark Bennet Nov 01 '21 at 16:44
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    For the ellipse and hyperbola you are using formulas that assume the center of the ellipse or hyperbola is at the origin. A non-degenerate ellipse or hyperbola never passes through its center. That's why you can't find a formula. Try a center somewhere else. – David K Nov 01 '21 at 16:58
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    For the parabola, did you try the hint? It says to rotate the three points first. You have to do that before you apply the formula. – David K Nov 01 '21 at 16:59
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    If you still need better answers, try writing out the equations in MathJax so people can read them better; your equations are easy to write that way and you should find all you need to get started here: https://math.stackexchange.com/help/notation ... And then you can fill in some sentences between equations explaining what you were thinking and why you decided you needed that equation. – David K Nov 01 '21 at 17:02
  • In Euclidean Geometry, there exists a circle between any 3 non-collinear points. For the circle, the perpendicular bisector of any two chords intersects at the center. For the parabola, you need to rotate the starting points so the parabola passes the vertical line test, then do quadratic polynomial fit. – TurlocTheRed Nov 01 '21 at 17:42

2 Answers2

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Construct a family of conics with intercepts $(0,0)$, $(1,0)$ and $(0,1)$:

$$ax^2+2hxy+by^2-ax-by=0$$

$$\Delta=\det \begin{pmatrix} a & h & -\frac{a}{2} \\ h & b & -\frac{b}{2} \\ -\frac{a}{2} & -\frac{b}{2} & 0 \\ \end{pmatrix} =\frac{ab(2h-a-b)}{4} \\$$

  • $ab-h^2>0 \implies$ ellipse (always real)

  • $ab-h^2=0$ and $\Delta\ne 0 \implies$ parabola

  • $ab-h^2<0$ and $\Delta\ne 0 \implies$ hyperbola

  • $a=0 \implies y(2hx+by-b)=0$, two straight lines

  • $b=0 \implies x(ax+2hy-a)=0$, two straight lines

  • $h=\frac{a+b}{2} \implies (x+y-1)(ax+by)=0$, two straight lines

See another post of mine for your interests.

Ng Chung Tak
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General conic: $g(x,y)=Ax^2+Bx+Cy^2+Dy+Exy+F=0$

$g(0,0)=0\implies F=0.$

$g(0,1)=0 \implies C+D=0\implies D=-C$

$g(1,0)=0 \implies A+B=0 \implies B=-A$

So: $g(x,y)=A(x-1/2)^2+C(y-1/2)^2+Exy -(A+C)/4=0$

Any choice of A,C, and E will give you a conic with the three mentioned points on it. Let $E=0$ and $A=C=1$, we have a selection of coefficients that produces a circle.

More generally, using the quadratic formula to solve for $y$ in terms of $x$:

$$y=\frac{(C-Ex)+\sqrt{(E^2-4AC)x^2+(4AC-2EC)x+C^2}}{2C}$$

Here the discriminant is a function of $x$ that represents a parabola with axis parallel to the y axis.

Given a parabola of the form $y=mx^2+nx+p$, it has a minimum if m is positive, a max if m is negative. In either case, the extremal value of y is $y_e=-n^2/4m+p$.

Properties of this parabola tell you about the conic. If it's a perfect square, the conic is a line. If its negative sufficiently far from the origin, the conic is an ellipse. Otherwise its a hyperbola or a parabola, largely depending on if it touches or crosses the x axis. Parabola if touching, hyperbola if its positive for all points above a certain distance from the origin.

TurlocTheRed
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