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## Hm, math help? Parabolas...

• Posted by a hidden member.
I've been out of school for a long time and I am scratching my head with this one because I need to recreate a parabola from a drawing for an architecture project.

The parabola passes through points (0,7), (x,0) and (6,2). I can't, for the life of me, figure out the equation of the parabola and the missing point, x. It is tangential to the x-axis and doesn't pass below it.

I need it so I can create it and then scale it up or down. It's a component for the base of a Corinthian column. Couldn't find one in any CAD libraries so I'm building my own from scratch. It's not as easy as measuring off of the drawing because, although the drawing has all of the proportions labelled, it isn't to scale itself.

I figured this might be the best forum to post under

Parabola: https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/179966_919508581219_94806804_40364904_467605504_n.jpg
• Posted by a hidden member.
Damn, that unknown in the coordinates is a real bitch!

What I would try is to just treat the x like a real number and insert the coordinates into the parabolic equation y = ax^2 + bx + c. Once you have the three equations, try isolating some variables and placing them into the other equations in order to simplify and cancel stuff out. What's gonna be tough though is when you work with the (x, 0) equation seeing how you end up with nothing but the parabolic formula itself.

Waiiiiittt, just a thought here but you said that the parabola is tangential to the x-axis? Well with the point (x, 0) it must mean that this point is the absolute minimum because y ≥ 0. So would that be the vertex?
• Posted by a hidden member.
Rawrly said
Lash saidWaiiiiittt, just a thought here but you said that the parabola is tangential to the x-axis? Well with the point (x, 0) it must mean that this point is the absolute minimum because y ≥ 0. So would that be the vertex?

Yes, y ≥ 0 !!!

I haven't worked with parabolas in while so don't laugh if I say something retarded

Anyways, so I'm thinking that the point (x, 0) is the vertex. Because of that, can we use the points (0,7) and (6,2) to find the equation of the parabola thus using the equation to find the vertex? Of course when you find the equation of the parabola it'll probably been in standard form so you're gonna have to switch it into vertex form.
• Posted by a hidden member.
Well like, for the first point (0, 7), we'd plug it into the equation Ax^2 + Bx + C = 0, giving us:

(0, 7): 0 + 0 + C = 7 -> C = 7

for the second,

(6, 2): 36A + 6B + C = 2

We can then combine them since we know that C = 7:

36A + 6B + 7 = 2
36A + 6B + 7 - 5 = 0
36A + 6B + 2 = 0

I dunno if that helped at all
• Posted by a hidden member.
Parabolas are stupid. All they do is make aspiring astronauts feel weightless for a few seconds at a time.
• Posted by a hidden member.
I don't speak Numerical.
• Posted by a hidden member.
JoyfullyRandom saidI don't speak Numerical.
Parabolas are not numerical. They're just a stupid anti-gravity loop that almost every pilot has done for fun.
• Posted by a hidden member.
I remember parabolas from my sophomore year in high school. Unfortunately I don't remember how to do them and will probably never use algebra of that level again in my life. The American education system is awesome
• Posted by a hidden member.
People. Linear algebra.

Know it. Use it.

You have two linear equations in three variables (a, b, c). You are asked to solve for the case where y = 0. The solution is actually an infinite range of parabolas that happen to intersect the y-axis at 7, constrained by the relationship:

a*x^2 + b*x = -7

The difference will be the rate of curvature of the parabola. You can come to a solution if you specify an additional linear equation acting as a constraint, say one that specified where the x-intercepts of the parabola were.
• Posted by a hidden member.
abelian0 saidPeople. Linear algebra.

Know it. Use it.

You have two linear equations in three variables (a, b, c). You are asked to solve for the case where y = 0. The solution is actually an infinite range of parabolas that happen to intersect the y-axis at 7, constrained by the relationship:

a*x^2 + b*x = -7

The difference will be the rate of curvature of the parabola. You can come to a solution if you specify an additional linear equation acting as a constraint, say one that specified where the x-intercepts of the parabola were.

I was thinking it had something to do with linear algebra! Way too terrifying for me to think of though.
• Posted by a hidden member.
Linear algebra is one of the most useful subjects anyone can study, ever. Superposition is a powerful concept, unless we're talking about my shower. For whatever reason, the nozzle water temperature is nonlinear in H and C, where H is the number of degrees the hot water knob is turned, and C is the number of degrees the cold water knob is turned.
• Posted by a hidden member.
No, I agree, linear is extremely useful. I just don't like it. Aced the course in first year and don't have to do it again, so it's all good. I'm a calculus guy ;)
• vintovka

Posts: 588
Not to nerd out too much but tecnically the parabola is the conic section taken by passing a plane through a cone parallel to the side, It's fundamentally a figure in Euclidean geometry. It cn be expressed in algebraic form, but it's really a geometric figure that is more ancient than algebra. (of course since Descartes everyone uses algebra for geometry, but geometry is older)

The above comment is not going to help anyone to draw a parabola, but is interesting as trivia (or maybe uninteresting as trivia).
• Posted by a hidden member.
Okay, I'll admit I loved multivariable/vector calculus more than even linear algebra. But I couldn't decide between topology and Galois theory if you put a gun to my head. Both very beautiful in entirely different ways, unless of course you get to algebraic geometry and that's a whole other topic (my advisor at Berkeley was Robin Hartshorne, who literally wrote the book on algebraic geometry).
• Posted by a hidden member.
My professor researches in the field of topology, remarkable stuff. I actually had some discussions with him about it and even though it's way out of my league I'm so stoked to learn it!
• Posted by a hidden member.
vintovka saidNot to nerd out too much but tecnically the parabola is the conic section taken by passing a plane through a cone parallel to the side, It's fundamentally a figure in Euclidean geometry. It cn be expressed in algebraic form, but it's really a geometric figure that is more ancient than algebra. (of course since Descartes everyone uses algebra for geometry, but geometry is older)

The above comment is not going to help anyone to draw a parabola, but is interesting as trivia (or maybe uninteresting as trivia).

Nah, I'm just playing It's actually some cool info
• Posted by a hidden member.
Good start Lash, and the fact that it is tangent to the x-axis provides the additional constraint so that there isn't an infinite range of possible parabolas.

I can try to think of a more elegant way, but this works and isn't too bad:

So we have 36a+6b+5=0 as above.

It's easy to see (by completing the square or taking the derivative) that the minimum of a parabola occurs at x= -b/(2a), so if we put that in for x we should get 0, i.e.

0 = -a(-b/2a)^2 + b(-b/2a) + 7 , which after simplifying and multiplying through by 4a (since a can't equal 0) tells us b^2=28a.

Two equations, two variables. Throw a = b^2/28 into first equation and solve the quadratic. You get b = (-7-sqrt14)/3 = -3.58055 so a = .45787 and thus the x in your picture is -b/2a = 3.92001

(The quadratic for b also gives you (-7+sqrt14) / 3 which also gives you a parabola through those points and tangent to the x-axis. However in this case, the point of tangency is to the right of (6,2).)
• Posted by a hidden member.
Draw a directrix (horizontal line) below the parabola at y = -F

Call the x-intercet (t, 0) so not to confuse the x's.

Draw the Focus at the point (t, F)

Definition of a parabola: the set of points equidistant from the focus and the directrix.

So the distance from the left point (0,7)
and the focus is (cartesian formula)
= Sqrt[ ( 0 - t )**2 + ( 7 - F )**2 ] [eq1]

The distance from that point to the directrix is 7 + F [eq2]

[eq1] = [eq2]

Similarly, the right point is

= Sqrt[ ( 6 - t )**2 + ( 2 - F )**2 ] from the Focus and [eq3]

2 + F from the directrix. [eq4]

[eq3] = [eq4]

Now... Solve for F in terms of t with these two equations.

Relook at the left point (0, 7) and do the distance again between the focus and the directrix now substituting F.

You'll get a value for t, then for F.

The general parabola formula is ( x - h )**2 = 4p( y - k )

where the directrix is y = -p
and (h, k) is the vertex... (t, 0) in our problem

So, in our case k = 0, and h = t, p = F

Plug those in - you have your equation.

• Posted by a hidden member.
Since the parabola doesn't go below zero, it can be expressed as:

y = a(x-b)2

We know that points (0,6) and (6,2) meet this equation, hence we know that:

6 = a(0-b)2
2 = a(6-b)2

From the former it follows that:

a = 6/b2

If we substitute a in the latter we get:

2 = 6/b2 * (6-b)2

Which can be simplified to:

b2 - 18b + 54 = 0

Then b = 9 + 3√3 or b = 9 - 3√3

y = 0 only when x = b. Then the b values we've found are the answer we were looking for!
• RunnerMD

Posts: 157
Math geeks on RJ. As a math/computer geek I love it!
• Posted by a hidden member.
Divide by zero, anyone?

• Posted by a hidden member.
djshowdown2 saidGood start Lash, and the fact that it is tangent to the x-axis provides the additional constraint so that there isn't an infinite range of possible parabolas.

I can try to think of a more elegant way, but this works and isn't too bad:

So we have 36a+6b+5=0 as above.

It's easy to see (by completing the square or taking the derivative) that the minimum of a parabola occurs at x= -b/(2a), so if we put that in for x we should get 0, i.e.

0 = -a(-b/2a)^2 + b(-b/2a) + 7 , which after simplifying and multiplying through by 4a (since a can't equal 0) tells us b^2=28a.

Two equations, two variables. Throw a = b^2/28 into first equation and solve the quadratic. You get b = (-7-sqrt14)/3 = -3.58055 so a = .45787 and thus the x in your picture is -b/2a = 3.92001

(The quadratic for b also gives you (-7+sqrt14) / 3 which also gives you a parabola through those points and tangent to the x-axis. However in this case, the point of tangency is to the right of (6,2).)

I completely blew over the part where he said it was tangent to the x-axis, hence why I made that statement about introducing some other constraint like x-intercepts (which clearly is the minimum in this case).

Good job spotting that oversight. It's what I get for not RTFM.
• Posted by a hidden member.
JPtheBITCH saidSpeaking of pay grades, I'm way beyond mine in this discussion. Calculus was 39 years ago.

I'm much more a fan of hyperbole.

Which RJ abounds in
• Posted by a hidden member.
Radamisto saidSince the parabola doesn't go below zero, it can be expressed as:

y = a(x-b)2

We know that points (0,6) and (6,2) meet this equation, hence we know that:

6 = a(0-b)2
2 = a(6-b)2

From the former it follows that:

a = 6/b2

If we substitute a in the latter we get:

2 = 6/b2 * (6-b)2

Which can be simplified to:

b2 - 18b + 54 = 0

Then b = 9 + 3√3 or b = 9 - 3√3

y = 0 only when x = b. Then the b values we've found are the answer we were looking for!

Best way to do this.
• Posted by a hidden member.