The arch of an underpass is a semi - ellipse 60ft wide and 20 ft high. find the clearance at the edge of a lane if the edge is 20ft from the middle.

Backofbook: $\displaystyle \frac{20}{3} \sqrt{5} ft$

:confused:

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- Feb 8th 2007, 01:35 AM^_^Engineer_Adam^_^Ellipse help
The arch of an underpass is a semi - ellipse 60ft wide and 20 ft high. find the clearance at the edge of a lane if the edge is 20ft from the middle.

Backofbook: $\displaystyle \frac{20}{3} \sqrt{5} ft$

:confused: - Feb 8th 2007, 02:23 AMticbol
Don't I love ellipses.

So the ellipse is horizontal. Its major axis, 2a, is 60 ft. Its minor axis, 2b, is 2*20 ft.

If the center of the ellipse is the center of the lane, then the ellipse is

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

(x^2)/(30^2) +(y^2)/(20^2) = 1 -----(i)

At the edge of the lane, or at x = 20 ft, what is the y there?

(20^2)/(30^2) +(y^2)/(20^2) = 1

Clear the fractions, multiply both sides by (30^2)(20^2),

(20^2)(20^2) +(y^2)(30^2) = (30^2)(20^2)

(30^2)(y^2) = (30^2)(20^2) -(20^2)(20^2)

(30^2)(y^2) = (20^2)(30^2 -20^2)

(30^2)(y^2) = (20^2)[(30+20)(30-20)]

(30^2)(y^2) = (20^2)(50*10)

(30^2)(y^2) = (20^2)(500)

(30^2)(y^2) = (20^2)(10^2)(5)

y^2 = [(20^2)(10^2)(5)] / (30^2)

Take the square roots of both sides,

y = [(20)(10)sqrt(5)] / 30

y = 200sqrt(5) / 30

y = (20/3)sqrt(5) ft. ---------------answer. - Feb 8th 2007, 04:40 AMSoroban
Hello, ^_^Engineer_Adam^_^!

ticbol provided an excellent solution . . . as usual.

. . I simplified the arithmetic/algebra.

Quote:

The arch of an underpass is a semi-ellipse, 60ft wide and 20 ft high.

Find the clearance at the edge of a lane if the edge is 20ft from the middle.

Back of book: $\displaystyle \frac{20}{3} \sqrt{5}$ ft

The equation is: .x²/30² + y²/20² .= .1

. . . . . . . . . . . . . . . . . . . _______

And we have: . y .= .(2/3)√900 - x²

. . . . . . . . . . . . . . . . . . . ___ . . . . . . . . ._ . . . . . . . . ._

When x = 20: .y .= .(2/3)√500 .= .(2/3)10√5 .= . (20/3)√5

I'm still waiting for LaTeX to return . . .