Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
End-of-Course Assessment

Exercise 57 Page 969

What is the standard equation of a horizontal ellipse?

B

Practice makes perfect

We have a bridge in the shape of an arch that looks like the upper half of an ellipse. This bridge spans a distance of 80 feet, and the maximum height of the bridge is 30 feet. Let's draw a horizontal ellipse to represent the bridge geometrically on a coordinate plane.

We want to find the height of the arch 28 feet from the center. To do so, we first need to write an equation to represent this ellipse. Let's remember the standard equation of a horizontal ellipse. x^2/a^2+y^2/b^2=1, a> b>0 In this form a represents the vertices of the ellipse and b represents the co-vertices. For our ellipse, we will substitute the positive values of vertices, a= 40 and b= 30, into the equation and then simplify it.

x^2/a^2+y^2/b^2=1
x^2/40^2+y^2/30^2=1
x^2/1600+y^2/900=1

Now, we will substitute x= 28 into our equation and solve it for y to find the height of the arch in that point.

x^2/1600+y^2/900=1
28^2/1600+y^2/900=1
â–¼
Solve for y
784/1600+y^2/900=1
784/1600+y^2/900=1600/1600
y^2/900=1600/1600-784/1600
y^2/900=816/1600
y^2=816/1600 * 900
y= ± sqrt(816/1600 * 900)
y= ± 21.424285 ...
y ≈ ± 21.4

Since a negative height does not make sense, the answer should be positive. Therefore, the height of the arch 28 feet from the center is about 21.4 feet. which corresponds to option B.