Question

HELP PLEASE

A railroad tunnel is shaped like a semi ellipse, as shown below.



The height of the tunnel at the center is 27 ft, and the vertical clearance must be 9 ft at a point 24 ft from the center. Find an equation for the ellipse.

Answer 1

General equation for an ellipse with the center (0,0) could be determined using

`\boxed{ (x^(2))/(a^(2))+ (y^(2))/(b^(2))=1}`
with stands for the length of the horizontal radius and b stands for the length of vertical radius

From the question above, we could draw informations:
b = 27 ft
one of the points on ellipse (x,y) = (24,9)

plug the numbers into the equation

`(x^(2))/(a^(2))+ (y^(2))/(b^(2))=1`

`(24^(2))/(a^(2))+ (9^(2))/(27^(2))=1`

`(576)/(a^(2))+ (81)/(729)=1`

`(576)/(a^(2))+ (1)/(9)=1`
solve for a²

`(576)/(a^(2)) =1-(1)/(9)`

`(576)/(a^(2)) =(8)/(9)`

`a^(2) =(9)/(8) * 576`
a² = 648

Input the value of a² to the equation

`\boxed{ (x^(2))/(648)+ (y^(2))/(729)=1}`
This is the equation of the ellipse

Answer 2

Remark
The general equation for an ellipse is

`(x^2)/(a^2)+ (y^2)/(b^2)=1`

At the center, the the point is 0,27

Solving for the

`(0)/(a^2) + (27^2)/(b^2) = 1`
or b^2= 27^2 * 1
Taking the square root of both sides
b = 27

Now we need to use one more piece of information
The second point is (24,9) what that means is at the point x = 24, y = 9

Solving for that

`(24^2)/(a^2)+(9^2)/(27^2)=1`

576/a^2 + 81/729 = 1 Subtract 81/729 from both sides
576/a^2 = 1 - 81/729 = 576/a^2 = (729 - 81)/729 = 648 /729

576/a^2 = 648/729 Cross multiply

648 a^2 = 729 * 576 Divide by 648
a^2 = 648

Full equation for the ellipse and answer is


`(x^2)/(648)+ (y^2)/(729) = 1`