Find a point on the parabola with given point and distance

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To find a point on the parabola y = x^2 that is 4*sqrt(17) units away from the point (18,0), the distance formula is applied. The equation simplifies to 272 = y^2 + (x - 18)^2, leading to y^2 = -(x - 18)^2 + 272. A quartic equation emerges when substituting y^2 with x^4, allowing for the identification of roots. The discussion emphasizes the importance of rewriting equations to isolate variables for easier solving. Ultimately, the problem involves finding a specific point on the parabola based on the given distance criteria.
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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.
 
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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:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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