Answer: a) [tex]\frac{x^{2} }{2352.25 }[/tex] + [tex]\frac{y^{2} }{529}[/tex] = 1
b) The distance of two foci is 85.4 feet
c) Area = 3502.67 square feet
Step-by-step explanation: a) An ellipse has the equation in the form of:
[tex]\frac{x^{2} }{a^{2} }[/tex]+[tex]\frac{y^{2} }{b^{2} }[/tex] = 1, where a is the horizontal axis and b is the vertical axis.
For the Statuary Hall, a = [tex]\frac{97}{2}[/tex] = 48.5 and b = [tex]\frac{46}{2}[/tex] = 23, so the equation will be
[tex]\frac{x^{2} }{2352.25 }[/tex] + [tex]\frac{y^{2} }{529}[/tex] = 1.
b) To determine the distance of the foci, we have to calculate 2c, where c is the distance between one focus and the center of the ellipse. To find c, as a, b and c create a triangle with a as hypotenuse:
[tex]a^{2}[/tex] = [tex]b^{2} + c^{2}[/tex]
[tex]c^{2} = a^{2} - b^{2}[/tex]
c = [tex]\sqrt{48.5^{2} - 23^{2} }[/tex]
c = 42.7
The distance is 2c, so 2·42.7 = 85.4 feet.
The two foci are 85.4 feet apart.
c)The area of an ellipse is given by:
A = a.b.π
A = 48.5 · 23 · 3.14
A = 3502.67 ft²
The area of the floor room is 3502.67ft².