Find a point on the parabola with given point and distance

  • Thread starter Thread starter theintarnets
  • Start date Start date
  • Tags Tags
    Parabola Point
Click For Summary

Homework Help Overview

The problem involves finding a point on the parabola defined by the equation y = x², such that the distance from this point to the point (18, 0) is equal to 4√17. The original poster expresses uncertainty about how to apply the distance formula and how to proceed with their calculations.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the distance formula but is unsure about their substitutions and the next steps. Some participants suggest rewriting the equations to express one variable in terms of another, leading to a quartic equation. There are questions about the meaning of finding roots and the reasoning behind substituting y² with x⁴.

Discussion Status

The discussion is ongoing, with some participants providing guidance on how to approach the problem. There is an acknowledgment of different interpretations regarding the nature of the distance being sought. The original poster expresses confusion but also indicates they have made progress independently.

Contextual Notes

There is a mention of a potential misunderstanding regarding the nature of the problem, as one participant notes a distinction between finding a point at the shortest distance versus a specific distance. The original poster's uncertainty about mathematical concepts and terminology is evident.

theintarnets
Messages
64
Reaction score
0

Homework Statement



Find a point on the parabola y = x^2 whose distance from (18,0) is 4*sqrt(17)

Homework Equations



I'm guessing I need to use the distance formula, but I'm not entirely sure.
D = sqrt( (x2 - x1)^2 + (y2 - y1)^2)

The Attempt at a Solution



Well I really don't think I'm doing this right. I tried substituting the given point and distance into the equation, but I'm not really sure where to go from there.

4*sqrt(17) = sqrt( (18 - x)^2 + (0 - y)^2)
272 = y^2 + x^2 - 36x + 324
y^2 = -(x-18)^2 + 272

I'm not really sure what to do. Any help would be greatly appreciated.
 
Physics news on Phys.org
welcome to pf!

hi theintarnets! welcome to pf! :smile:

that's ok so far …

now put y2 = x4

that gives you a quartic equation …

but you should be able to spot one root by guesswork (you only need one :wink:)​
 
I'm sorry, I'm not very good at math and I have no idea what you mean.
So I substitute x^4 for y^2? Can you explain why? And I'm not 100% sure what you mean by finding the roots. I'm so ashamed :cry:
 
hi theintarnets! :smile:

(just got up :zzz: …)
theintarnets said:

hmm … that's a different problem, finding a point at the shortest distance rather than at an actual distance
theintarnets said:
I'm sorry, I'm not very good at math and I have no idea what you mean.
So I substitute x^4 for y^2? Can you explain why?

this is a standard method for solving a pair of simultaneous equations (ie two equations with two unknowns, x and y) …

you rewrite one equation to express y in terms of x (or vice versa),

then you put that value of y into the other equation, giving you an equation with only one unknown (x)! :smile:
… And I'm not 100% sure what you mean by finding the roots.

the solutions of a polynomial equation are also known as the roots …

in this case, there's an obvious whole number that works … try it! :wink:
 

Similar threads

Replies
19
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
9K