An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 6 feet and a span of 22 feet. Find an equation for the ellipse. (Assume a center at (0, 0)). Use the equation to find the height of the arch at a distance of 3 feet from the center. (Round your answer to two decimal places.)

Answer :

The equation of ellipse is [tex]\frac{x^2}{11^2} + \frac{y^2}{6^2} = 1[/tex] and height of arch is 5.77 feet

Assume that the center of the ellipse is at the origin.

So,  The general equation for an ellipse is:

[tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]

Where:

x is Horizontal distance, in feet.

y is Vertical distance, in feet.

a is Horizontal semiaxis length (half-width), in feet.

and b is Vertical semiaxis length (height), in feet.

According to the question,

x = 3 , a = 11 and b = 6

Putting all the values ,

So , the equation of ellipse will be

[tex]\frac{x^2}{11^2} + \frac{y^2}{6^2} = 1[/tex]

Putting x = 3,

[tex]\frac{3^2}{11^2} + \frac{y^2}{6^2} = 1[/tex]

Solving for y,

[tex]\frac{y^2}{36} = 1 - \frac{9}{121}[/tex]

=> [tex]y^2 = \frac{(121 - 9)36}{121}[/tex]

=> y² = 4032 / 121

=> y = √33.32

=> y = 5.77 feet

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