Arch involving CONICS hyperbola equation

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Homework Help Overview

The original poster is tasked with designing a curved arch modeled by a hyperbola equation, specifically focusing on a horizontal span of 100 m and a maximum height of 20 m. They are attempting to find the parameters of the hyperbola equation given certain constraints.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the determination of the center (h, k) and the parameter a for the hyperbola. There are attempts to manipulate the equation and questions about expanding expressions and solving quadratics.

Discussion Status

There is an ongoing exploration of the quadratic equation derived from the hyperbola setup. Some participants provide guidance on solving for k and using the quadratic formula, while others express confusion about the signs and the implications of the hyperbola's equation.

Contextual Notes

Participants are working within the constraints of the problem, including the specific form of the hyperbola and the values derived from the arch's dimensions. There is mention of conflicting interpretations regarding the signs in the equations and the resulting calculations.

aisha
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Your task is to design a curved arch similar to the a tunnel for cars. with a horizontal span of 100 m and a maximum height of 20 m.

Using a domain of {x:-50<=x<=50} and {y:0<=y<=20} determine the following types of equations that could be used to model the curved arch.

the equation of a hyperbola in the form [tex]\frac {(x-h)^2} {a^2} - \frac {(y-k)^2} {b^2} = -1[/tex] where b=10 and the lower arm of the hyperbola would represent the arch.

How do i find a and the center (h,k) please help me I am struggling on this question.
 
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i got h=0 and y=20 because the coordinate (0,20) is on the y-axis

a cancels out because 0/a^2 =0 so only left with k as unknown get a quadratic function and get two values for k, which value do I take? choose a point let's say (0,50) and find a, these are the steps i have so far.

Um for k -(20-k)^2 how do u expand this (-20+k) (-20+k) or (-20+k) (20-k)?
 
I didn't follow your problem,but i can tell u for sure that

[tex]-(20-k)^{2}=-(20-k)(20-k)=(20-k)(k-20)[/tex]

Daniel.
 
ok I get [tex]\frac {k^2 +40 k-400} {100} =-1[/tex] is it possible to solve for k? once I find k i can plug it into the original equation to get a and then my standard form equation for the hyperbola representing the arch will be complete.
 
Sure you can solve for k. It's a simple quadratic. Multiply both sides by 100 and then use the quadratic formula.
 
hold on i changed what I did

[tex]-\frac {(20-k)^2} {100} = -1[/tex]

ok I cross multiplied and got

[tex]400-40k+k^2=-100[/tex]
[tex]k^2 -40k+500=0[/tex]

this is my quadratic using the quadratic formula i keep getting a negative under the square root why? [tex]b^2-4(a)(c)[/tex] sqrt(-400)
teacher said that k=10 and 30 but I don't know how.
 
U can't get a negative under the square root.U should get 400.

Daniel.

P.S.It's +100 in the RHS.
 
it can't be +100 on the rhs because this is the equation of a hyperbola. Using the information givin I am trying to find the equation of the hyperbola that could represent a arch with a span of 100 metres and maximum height of 20 metres. I was trying to solve k so that I could sub this value into the original equation and then get a then my equation in standard form for the hyperbola will be complete.
 
Honey,from

[tex]-\frac{(20-k)^{2}}{100}=-1[/tex]

please trust me that it follows

[tex]\frac{(20-k)^{2}}{100}=1 \Rightarrow (20-k)^{2}=100 \Rightarrow k_{1}=10,k_{2}=30[/tex]

,okay,sweetheart?


Daniel.
 
  • #10
the 100 is positive because you did what? :cry:

are we cross multiplying? or multiplying both sides by -100? what are we doing?

That equation i wrote has a negative sign too did u see that? before the brackets.
 
  • #11
I simplified an equality through "-1".Or,if u prefer,i multiplied both sides through the same skinny "-1".

Daniel.
 
  • #12
dextercioby said:
Honey,from

[tex]-\frac{(20-k)^{2}}{100}=-1[/tex]

please trust me that it follows

[tex]\frac{(20-k)^{2}}{100}=1 \Rightarrow (20-k)^{2}=100 \Rightarrow k_{1}=10,k_{2}=30[/tex]

,okay,sweetheart?


Daniel.

Leave this for your spouse, please.
 

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