Question 199175
A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular coordinate system and find the equation of the parabola. Then calculate the height of the arch 10, 20 and 40 feet from the center. Provide a sketch representing the situation and be sure it is clearly labled with coordinates.

{{{drawing(400,400,-10,130,-10,60, locate(40,30,Vertex(60,25)),

graph(400,400,-10,130,-10,60,(sqrt(x)/sqrt(x))*(sqrt(120-x)/sqrt(120-x))*(-1/144)(x-60)^2+25),line(60,24.8,60,25.2)


)
 )}}}

The equation of the parabola with vertex (h,k) is

{{{y=a(x-h)^2+k}}}

So the equation of the parabola with vertex (60,25) is

{{{y=a(x-60)^2+25}}}

Since it goes through (0,0) we substitute that in:

{{{0=a(0-60)^2+25}}}

{{{0=a(-60)^2+25}}}

{{{0=a(3600)+25}}}

{{{0=3600a+25}}}

{{{-3600a=25}}}

{{{a=25/(-3600)}}}

{{{a=-1/144}}}

So the equation of the parabola is

{{{y=(-1/144)(x-60)^2+25}}}

To find the height 10 feet from the center,
since the center is at 60, 10 feet from the
center is where x=50 and where x=70
we can either substitute {{{x=50}}} or {{{x=70}}}

Substituting 50,

{{{y=(-1/144)(50-60)^2+25}}} 

{{{y=(-1/144)(-10)^2+25}}}

{{{y=(-1/144)(100)+25}}}

{{{y=24.3056}}} feet, approximately

{{{drawing(400,400,-10,130,-10,60, locate(40,30,Vertex(60,25)),
line(50,0,50,875/36), line(70,0,70,875/36),
graph(400,400,-10,130,-10,60,(sqrt(x)/sqrt(x))*(sqrt(120-x)/sqrt(120-x))*(-1/144)(x-60)^2+25),line(60,24.8,60,25.2)) 
 )}}}

To find the height 20 feet from the center,
since the center is at 60, 20 feet from the
center is where x=40 and where x=80
we can either substitute {{{x=40}}} or {{{x=80}}}

Substituting 40,

{{{y=(-1/144)(40-60)^2+25}}} 

{{{y=(-1/144)(-20)^2+25}}}

{{{y=(-1/144)(400)+25}}}

{{{y=22.2222}}} feet, approximately

{{{drawing(400,400,-10,130,-10,60, locate(40,30,Vertex(60,25)),
line(40,0,40,200/9), line(80,0,80,200/9), 
graph(400,400,-10,130,-10,60,(sqrt(x)/sqrt(x))*(sqrt(120-x)/sqrt(120-x))*(-1/144)(x-60)^2+25),line(60,24.8,60,25.2)) 
 )}}}

To find the height 40 feet from the center,
since the center is at 60, 40 feet from the
center is where x=20 and where x=100
we can either substitute {{{x=20}}} or {{{x=100}}}

{{{y=(-1/144)(20-60)^2+25}}} 

{{{y=(-1/144)(-40)^2+25}}}

{{{y=(-1/144)(1600)+25}}}

{{{y=13.6111}}}feet, approximately

{{{drawing(400,400,-10,130,-10,60, locate(40,30,Vertex(60,25)),
line(20,0,20,245/18),line(100,0,100,245/18),
graph(400,400,-10,130,-10,60,(sqrt(x)/sqrt(x))*(sqrt(120-x)/sqrt(120-x))*(-1/144)(x-60)^2+25),line(60,24.8,60,25.2)) 
 )}}}


Edwin</pre>