Answer to Question #263932 in Analytic Geometry for Lovwly

Question

A cable hangs in a parabolic arc between two poles 100 feet apart. The poles are 30 feet high and the lowest point on a suspended cable is 5 feet above the ground.

a) find the equation of the arc if the vertex is the lowest point of the cable

b) the opening of a cave 20 feet wide is in the shape of a semi ellipse. If the height of the opening is 15 feet at the center, how high is the opening 5 feet from the center.

Answer 1

a) Let point $(0, 5)$ be the vertex of the parabola. Then the equation of the parabola will be

$$y=ax^2+5, a>0$$

Given $y(-50)=y(50)=30.$ Substitute

$$a(50)^2+5=30$$

$$a=0.01$$

The equation of the arc is

$$y=0.01x^2+5$$

b)

The equation of the ellipse is

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

Substitute

$$\dfrac{(x-0)^2}{(20/2)^2}+\dfrac{(y-0)^2}{15^2}=1$$

The equation of the given semi ellipse is

$$\dfrac{x^2}{100}+\dfrac{y^2}{225}=1, y\geq0$$

If $x_1=5$

$$\dfrac{(5)^2}{100}+\dfrac{y^2}{225}=1, y\geq0$$

$$y=\sqrt{225(1-1/4)}$$

$$y=7.5\sqrt{3}$$

The opening 5 feet from the center is $7.5\sqrt{3}(\approx13)$ feet high.