Question 632284
An arch in the form of a semiellipse is 64 ft wide at the base and has a height of 30 ft.
How high is the arch at a width of 20 ft from the center of the base? Round your answer
to the nearest tenth.
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Given problem is finding the equation of an ellipse with horizontal major axis and center at (0,0), then plugging in given x to find y.
Its standard form: {{{x^2/a^2+y^2/b^2+1}}}, a>b
a=32 ft (horizontal distance from center to vertex)
a^2=32^2=1024
b=30 ft (vertical distance from center to top of the arch)
b^2=900
equation: {{{x^2/1024+y^2/900=1}}}
solve for y when x=20 ft from center
..
{{{20^2/1024+y^2/900=1}}}
y^2/900=1-20^2/1024=1-400/1024=1-.391=0.609
y^2=.609*900=548.44
y=√548.44
y≈23.4
Height of the arch at a width of 20 ft from the center of the base≈23.4 ft