Find the equation of a parabola passing through 3 points

We looking for equation of parabola in standard form y = ax2+ bx+ c. Point (0,2) is point of parabola and because 2 = a•02 + b•0 + c, c = 2;

now y = ax2+ bx +2.

Points (3, 10) and (30, -8) give us system of two equations for a and b.

9a + 3b + 2 = 10,

900a + 30b + 2 = -8.

9a + 3b = 8

900a + 30b = -10

Let multiply first equation by -10 we and adding:

-90a - 30b = -80

900a + 30b = -10

810a = -90

a = -1/9, 9•(-1/9) + 3b = 8, -1 + 3b = 8, 3b = 9, b = 3.

Equatiopn of parabola y = -1/9x2 + 3x + 2

$\begingroup$

Good evening. How would I find the parametric equation of a parabola given three points? My point are endpoints $(-1, 2)$ and $(3, 4)$. However, the curve must also pass through $(0,0)$.

I have tried to solve in various ways using $x = at^2$ and $y = 2at$ but I keep going around in circles when I solve. I am really stumped.

Thank you so much in advance for your help.

Gabrielle

asked Jan 14, 2017 at 22:47

$\endgroup$

3

$\begingroup$

There are infinte parabolas passing through three points. But none of these have the $y$-axis as an axis.

The general equation of a parabola is $(Ax + By)^2+Cx+Dy+E=0$. Substituting $(-1,2)$, $(0,0)$, $(3,4)$, you can get any of the infinite number of solutions.

answered Jan 14, 2017 at 23:00

Sz_ZSz_Z

1,1206 silver badges11 bronze badges

$\endgroup$

2

$\begingroup$

There are an infinite number of parabolas passing through these three points, so I’ll proceed assuming that the parabola’s axis is meant to be parallel to the $y$-axis.

Plug your points into the general equation $y=ax^2+bx+c$ one by one and solve for the unknown coefficients. So, you have $$a(-1)^2+b(-1)+c=a-b+c=2,$$ which means $c=2-a+b$. Next, you have $$a(0)^2+b(0)+c=2-a+b=0,$$ from which $b=a-2$ and thus $c=0$. Finally, $$a(3)^2+b(3)+c=9a+3(a-2)+0=12a-6=4$$ which yields $a=5/6$.

I expect that you’ll be able to come up with some parameterization from this Cartesian equation on your own.

answered Jan 15, 2017 at 1:21

amdamd

51.3k3 gold badges29 silver badges78 bronze badges

$\endgroup$

0

There are two standard forms for a parabola:

Type 1 opens up or down:

#y = ax^2 + bx + c#

Type 2 opens left or right:

#x = ay^2 + by + c#

Plot the 3 points:

The points look like they fit the first type.

Write 3 equations by substituting the 3 points into the type 1 form:

#1 = a(1)^2 + b(1) + c" [1]"#
#-3 = a(-1)^2 + b(-1) + c" [2]"#
#1 = a(-3)^2 + b(-3) + c" [3]"#

Move the coefficients in front of the variables:

#1 = a + b + c" [1]"#
#-3 = a - b + c" [2]"#
#1 = 9a - 3b + c" [3]"#

Subtract equation [1] from equation [2]

#1 = a + b + c" [1]"#
#-4 =-2b" [2]"#
#1 = 9a - 3b + c" [3]"#

#1 = a + b + c" [1]"#
#2 =b" [2]"#
#1 = 9a - 3b + c" [3]"#

Substitute 2 for b into equations [1] and [3]:

#1 = a + 2 + c" [1]"#
#1 = 9a - 3(2) + c" [3]"#

#-1 = a + c" [1]"#
#7 = 9a + c" [3]"#

Subtract equation [1] from equation [3]:

#-1 = a + c" [1]"#
#8 = 8a" [3]"#

#-1 = a + c" [1]"#
#1 = a" [3]"#

Substitute 1 for a in equation [1]:

#c = -2#

Confirm that the equation: #y = x^2 + 2x - 2# fits all 3 points:

#1 = (1)^2 + 2(1) - 2#
#-3 = (-1)^2 + 2(-2) - 2#
#1 = (-3)^2 + 2(-3) - 2#

#1 = 1#
#-3 = -3#
#1 = 1#

This checks.

The equation is #y = x^2 + 2x - 2#

A parabola is given by the equation #y=ax^2+bx+c# which means that if the three coefficients #a#, #b# and #c# are known, the parabola is uniquely identified. Each point gives you a condition, and so, given three points you'll end up with three conditions for three variables, and thus there will be one solution, or no solutions at all.

To solve the problem, simply start from the generic formula written above, and substitute the points given. In fact, if a point #(x_0,y_0)# lies on the parabola, it must verify that #y_0=ax_0^2+bx_0+c#. So, with three given points #(x_1,y_1)##(x_2,y_2)# and #(x_3,y_3)# , you'll have the following system:
#y_1-ax_1^2-bx_1-c=0#
#y_2-ax_2^2-bx_2-c=0#
#y_3-ax_3^2-bx_3-c=0#

Remember that the variables are #a#, #b# and #c#. If you can solve this sistem, you can find the parabola.

Let me give you a simple example: pick the points #(0,0), (1,1)# and #(-1,1)#, belonging to the parabola #y=x^2#.
If #(0,0)# must lie on the parabola, then it must be
#0=a(0^2)+b(0)+c#, and thus #c=0#.
If #(1,1)# must lie on the parabola, then it must be
#1=a(1^2)+b(1)#, and thus #a+b=1#
If #(-1,1)# must lie on the parabola, then it must be
#1=a(-1^2)+b(-1)#, and thus #a-b=1#

So far, we know that #c=0# and that #a+b=1#, and #a-b=1#, and one can easily check that this means #a=1# and #b=0#.

We have thus solved the problem, because the solution #(a,b,c)=(1,0,0)# identifies the parabola #y=x^2#

Can you define a parabola with 3 points?

How do you find the equation for a parabola?