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2. SIGNS A sign is in the shape of an ellipse. The eccentricity is 0.60 and the length is 48 inches.

a. Write an equation for the ellipse if the center of the sign is at the origin and the major axis is horizontal.
b. What is the maximum height of the sign?

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a. The standard equation for an ellipse with center at the origin and major axis horizontal is:

x^2/a^2 + y^2/b^2 = 1

where a is the length of the semi-major axis and b is the length of the semi-minor axis. The eccentricity e is related to a and b by the equation:

e = √(a^2 - b^2)/a

We are given that the eccentricity e is 0.60 and the length of the major axis is 48 inches. Since the major axis is horizontal, a is half of the length of the major axis, so a = 24. We can solve for b using the equation for eccentricity:

0.6 = √(24^2 - b^2)/24

0.6 * 24 = √(24^2 - b^2)

14.4^2 = 24^2 - b^2

b^2 = 24^2 - 14.4^2

b ≈ 16.44

Therefore, the equation of the ellipse is:

x^2/24^2 + y^2/16.44^2 = 1

b. To find the maximum height of the sign, we need to find the length of the semi-minor axis, which is the distance from the center of the ellipse to the top or bottom edge of the sign. We can use the equation for the ellipse to solve for y when x = 0:

0^2/24^2 + y^2/16.44^2 = 1

y^2 = 16.44^2 - 16.44^2 * (0/24)^2

y ≈ 13.26

Therefore, the maximum height of the sign is approximately 26.52 inches (twice the length of the semi-minor axis).

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