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I need to find equations for parabolas which are passing through $(0,-1)$,$(-2,0)$,$(0,3)$,$(4,0)$. Help me. I know that parabolas determined by those four points but how to find equations explicitly?

JohnColtraneisJC
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friendlyuser
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  • What have you tried so far? Do you know the general equation for a parabola? – saulspatz Oct 01 '19 at 14:53
  • @saulspatz $y^2=2px $or$ y=ax^2+bx+c$. But im trying to solve this problem using projective Geometry and equation for conics – friendlyuser Oct 01 '19 at 14:55
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    You should edit your question to explain what you have tried and where your difficulty lies. – saulspatz Oct 01 '19 at 14:56
  • @friendlyuser: $y^2=2px$ and $y=ax^2+bx+c$ only give parabolas whose axes align with the coordinate axes. That cannot be the case here, since each coordinate axis contains two points of the target parabola. Do you know the general-general equation of a conic? Have you applied the techniques shown in answers to your previous question about parabolas? – Blue Oct 01 '19 at 15:09
  • There might not be a parabola for every set of 4 points. – Sam Oct 01 '19 at 15:12
  • @Sam: Confirmed. There are two solutions here. – Blue Oct 01 '19 at 15:13
  • @Blue you said that i cant use $y^2=2px$ and other one – friendlyuser Oct 01 '19 at 15:14
  • That's because that is the form a parabola with vertex at (0,0) which your's is clearly not. – Sam Oct 01 '19 at 15:16
  • @friendlyuser: Correct, you cannot use $y^2=2px$ or $y=ax^2+bx+c$. The parabolas here are not coordinate-axis-aligned. (The ones in your previous question weren't either, so this shouldn't be an entirely new phenomenon for you.) – Blue Oct 01 '19 at 15:16
  • You will have to use the generalized equation of a second degree conic. – Sam Oct 01 '19 at 15:18

1 Answers1

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Use the general parabola equation below,

$$ax^2 + 2\sqrt{ab}xy + b y^2 + gx + fy + c = 0$$

Plug in the points (0,-1) and (0,3) and, without losing generality, let $b=1$ for now,

$$1-f+c, \>\>\>9+3f+c = 0$$

which gives $f=-2$ and $c=-3$. Then, plug in the points (-2,0) and (4,0),

$$4a -2g-3=0,\>\>\>16a+4g-3=0$$

which gives $a=\frac 38$ and $g=-\frac 32$.

Thus, the equation for the parabola is,

$$\frac 38 x^2 + \sqrt{\frac 32 }xy + y^2 -\frac 32 x - 2y -3= 0$$

or, with $b=8$,

$$3 x^2 + 4\sqrt{6}xy + 8y^2 -12 x - 16y -24= 0$$

Edit: As pointed out by @Blue below. A second equation can be obtained by reserving the sign of the cross term.

Quanto
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