# Algebra II

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I'm confused..

1. The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface midway between the towers, what is the height of the cable at a point 150 feet from the middle of the bridge?

2. A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, how deep should the search light be?

What is be best to set up a graph for these? And what is a paraboloid?

• Algebra II -

1. The parabola shape is of the form y = a x^2, where y is the height and x is the horiziontal distance from the lowest point.

80 = a (300)^2
a = 80/(300)^2
When x = 150, y = 80*150^2/300^2 = 80*(1/4) = 20 feet

2. A paraboloid is the solid shape that you get when you rotate a parabola about its axis of symmetry. The light source will be located at the focus of the parabola so that the light that imerges is a parallel beam. (They should have told you that). Figure out the equation of the parabola from those facts and then calculate the depth of the parabola when the width is 5 feet.

• Algebra II -

Ah thanks!

• Algebra II -

The main cables of a suspension bridge uniformly distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 fi apart. The cables are attached to the towers at a height of 110 ft above the roadway and are 10 ft above the roadway at their lowest points. If vertical support cables are at 50-ft intervals along the level roadway, what are the lengths of these vertical cables?

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