Find an equation of the circle with the given center and radius.
-A bridge with a semi-elliptical arch extends over a river as shown. Find an equation of the ellipse that represents the arch of the bridge.

A) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 324 } = 1$
B) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 81 } = 1$
C) $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 144 } = 1$
D) $\frac { x ^ { 2 } } { 324 } + \frac { y ^ { 2 } } { 144 } = 1$

Question

Find an equation of the circle with the given center and radius.
-A bridge with a semi-elliptical arch extends over a river as shown. Find an equation of the ellipse that represents the arch of the bridge.

A) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 324 } = 1$ 
B) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 81 } = 1$ 
C) $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 144 } = 1$ 
D) $\frac { x ^ { 2 } } { 324 } + \frac { y ^ { 2 } } { 144 } = 1$

Quizplus · Accepted Answer

Since the arch is semi-elliptical, the equation of the arch is of the form:
$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
where $a$ and $b$ are the lengths of the semi-major and semi-minor axis, respectively.

We can find the values of $a$ and $b$ by looking at the center and endpoints of the arch. The center of the arch is at the origin $(0,0)$ and the endpoints are at $(-6,12)$ and $(6,12)$. The distance from the origin to either endpoint is 12, so the length of the semi-minor axis is $b=12$.

To find the length of the semi-major axis $a$, we can use the fact that the semi-ellipsis is half of the semi-elliptical arch.

The length of the semi-ellipse is given by:
$$l = \frac {\pi } { 2 } \sqrt { 2 { ( a ^ { 2 } + b ^ { 2 } ) } }$$

The length of the semi-elliptical arch is 12, so:
$$12 = \frac {\pi } { 2 } \sqrt { 2 { ( a ^ { 2 } + 12 ^ { 2 } ) } }$$
$$a ^ { 2 } + 144 = \frac {144\pi ^2 } { 8 }$$
$$a ^ { 2 } = \frac {36\pi ^2 } { 1 } - 144$$

Thus, the equation of the semi-elliptical arch is:
$$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 144 } = 1$$

Simplifying, we get:
$$\boxed {\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 144 } = 1}$$

Find an Equation of the Circle with the Given Center x2144+y2324=1\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 324 } = 1144x2​+324y2​=1

Find an Equation of the Circle with the Given Center $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 324 } = 1$