An arch for a bridge over a highway is in the form of a semiellipse. the top of the arch is 35 feet above ground (the major axis). what should the span of the bridge be (the length of its minor axis) if the height 25 feet from the center is to be 10 feet above ground?

For given problem:

Put midpoint of ellipse, (0,0) at epicenter of bridge at ground level.

Specified length of vertical major axis = 70=2a

a=35

a^2=1225

Equation of ellipse:

x^2/b^2+y^2=1

plug in coordinates of given point on ellipse(25, 10)

25^2/b^2 + 10^2/a^2 = 1

625/b^2 + 100/1225=1

625/b^2 = 1 - 100/ 1225 = .918

b^2 = 625/.918 ≈ 681

b ≈ 26.09

length of minor axis = 2b = 2(26.09) ≈ 52.16 ft

Span of bridge ≈ 52.16 ft

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3