anonymous
  • anonymous
A railroad tunnel is shaped like a semiellipse as shown below. The height of the tunnel at the center is 58 ft and the vertical clearance must be 29 ft at a point 21 ft from the center. Find an equation for the ellipse.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Diagram: http://prntscr.com/7mjbvf
jim_thompson5910
  • jim_thompson5910
not sure, but let me think
anonymous
  • anonymous
Anything?

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More answers

jim_thompson5910
  • jim_thompson5910
is there anything about the ellipse not mentioned? I feel like there's some missing info
anonymous
  • anonymous
nope
anonymous
  • anonymous
vertex is 0,58 and there are two points on the ellipse at +-21,29
jim_thompson5910
  • jim_thompson5910
if the ellipse is taller than it is wide, then a = 58 and this pairs up with the y^2 term
jim_thompson5910
  • jim_thompson5910
and if (21,29) is on the ellipse, then (x,y) = (21,29) --> x = 21 and y = 29
jim_thompson5910
  • jim_thompson5910
assume the center is (h,k) = (0,0)
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2} = 1\] \[\Large \frac{(21-0)^2}{b^2}+\frac{(29-0)^2}{58^2} = 1\] solve for b
anonymous
  • anonymous
24.25ish
jim_thompson5910
  • jim_thompson5910
actually you don't have to solve for b you can stop at b^2
anonymous
  • anonymous
well f***
jim_thompson5910
  • jim_thompson5910
since b^2 is in the denominator under the x^2 term
anonymous
  • anonymous
I'm getting a "Max Iterations Error" on my calculator
jim_thompson5910
  • jim_thompson5910
what kind of calculator do you have?
anonymous
  • anonymous
TI-36x Pro
jim_thompson5910
  • jim_thompson5910
I'm not familiar with that type
jim_thompson5910
  • jim_thompson5910
but you should get b^2 = 588
anonymous
  • anonymous
It has a number solve function where I can type in 2 sides of an equation and It'll solve for a variable in the equation. That's what I was using and it gave that error. When I solved for b^2 it works
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
hmm strange
anonymous
  • anonymous
ok so we found b^2 and b, so now what?
jim_thompson5910
  • jim_thompson5910
replace b^2 with 588 and a^2 with whatever 58^2 is
jim_thompson5910
  • jim_thompson5910
(h,k) = (0,0)
jim_thompson5910
  • jim_thompson5910
x and y are left alone in the equation
anonymous
  • anonymous
so \[\large \frac{ x^2 }{ 588 }+\frac{ y^2 }{ 3364 } =1\]
anonymous
  • anonymous
then we need to add the values
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
add values?
anonymous
  • anonymous
is that the equation or is there something left?
jim_thompson5910
  • jim_thompson5910
the last thing you posted is the equation they want
jim_thompson5910
  • jim_thompson5910
I guess you could solve for y to get some expression in the form y = ... that will get you the top half of the ellipse, which is the tunnel
anonymous
  • anonymous
should I do that?
jim_thompson5910
  • jim_thompson5910
hmm now I'm not sure if they want the full ellipse or just the upper half ellipse
anonymous
  • anonymous
ill give both
jim_thompson5910
  • jim_thompson5910
good idea
anonymous
  • anonymous
So do I just solve for y=?
jim_thompson5910
  • jim_thompson5910
yeah
anonymous
  • anonymous
So I'm stuck at \[\frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\]
jim_thompson5910
  • jim_thompson5910
multiply both sides by 3364 then take the square root of both sides you focus on the positive square root because you want the upper half
anonymous
  • anonymous
how I multiply x^2/588 by 3364 and what would that come out to
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\] \[\Large 3364*\frac{ y^2 }{ 3364 }=3364*(1-\frac{ x^2 }{ 588 })\] \[\Large y^2=3364*(1-\frac{ x^2 }{ 588 })\] \[\Large y^2=3364*1-3364*\frac{ x^2 }{ 588 }\] \[\Large y^2=3364-\frac{3364x^2 }{ 588 }\] \[\Large y^2=3364-\frac{841x^2 }{ 147 }\] \[\Large y=???\]
anonymous
  • anonymous
It was that last part I was confused about.\[y=58-\frac{ 29x }{ 7\sqrt{3} }\]
jim_thompson5910
  • jim_thompson5910
you can't take the square root like that
anonymous
  • anonymous
well crap
jim_thompson5910
  • jim_thompson5910
\[\Large y^2=3364-\frac{841x^2 }{ 147 }\] \[\Large \sqrt{y^2}=\sqrt{3364-\frac{841x^2 }{ 147 }}\] \[\Large y=\sqrt{3364-\frac{841x^2 }{ 147 }}\]
jim_thompson5910
  • jim_thompson5910
you apply the square root to the entire side
anonymous
  • anonymous
then what? Is that all we can do
jim_thompson5910
  • jim_thompson5910
yeah that's as far as you can go
anonymous
  • anonymous
Great thanks
jim_thompson5910
  • jim_thompson5910
\[\Large \sqrt{x + y} \ne \sqrt{x} + \sqrt{y}\]
anonymous
  • anonymous
It's been a long day of math for me. Finally off to bed. Thanks for all the help, dude!
jim_thompson5910
  • jim_thompson5910
no problem

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