Question

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.

Answer 1

Diagram: http://prntscr.com/7mjbvf

Answer 2

not sure, but let me think

Answer 3

Anything?

Answer 4

is there anything about the ellipse not mentioned? I feel like there's some missing info

Answer 5

nope

Answer 6

vertex is 0,58 and there are two points on the ellipse at +-21,29

Answer 7

if the ellipse is taller than it is wide, then a = 58 and this pairs up with the y^2 term

Answer 8

and if (21,29) is on the ellipse, then (x,y) = (21,29) --> x = 21 and y = 29

Answer 9

assume the center is (h,k) = (0,0)

Answer 10

\[\Large \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2} = 1\]

\[\Large \frac{(21-0)^2}{b^2}+\frac{(29-0)^2}{58^2} = 1\]

solve for b

Answer 11

24.25ish

Answer 12

actually you don't have to solve for b

you can stop at b^2

Answer 13

well f***

Answer 14

since b^2 is in the denominator under the x^2 term

Answer 15

I'm getting a "Max Iterations Error" on my calculator

Answer 16

what kind of calculator do you have?

Answer 17

TI-36x Pro

Answer 18

I'm not familiar with that type

Answer 19

but you should get b^2 = 588

Answer 20

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

Answer 21

ok

Answer 22

hmm strange

Answer 23

ok so we found b^2 and b, so now what?

Answer 24

replace b^2 with 588 and a^2 with whatever 58^2 is

Answer 25

(h,k) = (0,0)

Answer 26

x and y are left alone in the equation

Answer 27

so \[\large \frac{ x^2 }{ 588 }+\frac{ y^2 }{ 3364 } =1\]

Answer 28

then we need to add the values

Answer 29

@jim_thompson5910

Answer 30

add values?

Answer 31

is that the equation or is there something left?

Answer 32

the last thing you posted is the equation they want

Answer 33

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

Answer 34

should I do that?

Answer 35

hmm now I'm not sure if they want the full ellipse or just the upper half ellipse

Answer 36

ill give both

Answer 37

good idea

Answer 38

So do I just solve for y=?

Answer 39

yeah

Answer 40

So I'm stuck at \[\frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\]

Answer 41

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

Answer 42

how I multiply x^2/588 by 3364 and what would that come out to

Answer 43

\[\Large \frac{ y^2 }{ 3364 }=1-\frac{ x^2 }{ 588 }\]

\[\Large 3364*\frac{ y^2 }{ 3364 }=3364*(1-\frac{ x^2 }{ 588 })\]

\[\Large y^2=3364*(1-\frac{ x^2 }{ 588 })\]

\[\Large y^2=3364*1-3364*\frac{ x^2 }{ 588 }\]

\[\Large y^2=3364-\frac{3364x^2 }{ 588 }\]

\[\Large y^2=3364-\frac{841x^2 }{ 147 }\]

\[\Large y=???\]

Answer 44

It was that last part I was confused about.\[y=58-\frac{ 29x }{ 7\sqrt{3} }\]

Answer 45

you can't take the square root like that

Answer 46

well crap

Answer 47

\[\Large y^2=3364-\frac{841x^2 }{ 147 }\]

\[\Large \sqrt{y^2}=\sqrt{3364-\frac{841x^2 }{ 147 }}\]

\[\Large y=\sqrt{3364-\frac{841x^2 }{ 147 }}\]

Answer 48

you apply the square root to the entire side

Answer 49

then what? Is that all we can do

Answer 50

yeah that's as far as you can go

Answer 51

Great thanks

Answer 52

\[\Large \sqrt{x + y} \ne \sqrt{x} + \sqrt{y}\]

Answer 53

It's been a long day of math for me. Finally off to bed. Thanks for all the help, dude!

Answer 54

no problem