Word problems involving different conic sections

Question

1. A satellite dish has a shape called a paraboloid, where each cross-section is a parabola; since radio signals will bounce off the surface of the dish to the focus, the receiver should be placed at the focus. How far should the receiver be from the vertex, if the dish is 12ft across and 4.5ft deep at the vertex?
2. A street with only one lane is 15ft wide, goes through a semicircular tunnel with a diameter of 17ft. How high is the tunnel at the edge of the lane?
3. A tunnel has the shape of a semi-ellipse that is 15ft high at the center, and 36ft across the base. At most how high should a passing truck be, if it is 12ft wide, for it to be able to fit through the tunnel?

Answer 1

For the first question, we should take the information we were given to turn the parabola into the form

y = a(x-h)2 +k. As the problem doesn't place the parabola at a specific location in a coordinate grid, we can assume the vertex is at (0,0). Thus the equation simplifies to y = ax². We can solve for a given the information in the problem. Visualize the parabola as symmetric about the y-axis and you will see that we know that the parabola passes through (6, 4.5) and (-6, 4.5). These two points are 12 feet apart (wide) and 4.5 vertical feet away from the vertex (deep). Plugging 6 for y and 4.5 for x and solving for a, we see that a is 1/8. We can now use the formula for the focus f = (h, k+ 1/(4a)). Again, since we can assume the parabola is centered at (0,0), h and k are both 0. So, f = (0, 0+ 1/(4/8)) or (0,2). The receiver should be 2 feet away from the vertex.

The second question is much simpler than the initial information suggests. We can construct a right triangle to model the situation. Assuming the center of the lane is at (0,0), we know the right edge of the lane is at (7.5,0) since the lane is 15 feet wide. The becomes the horizontal side of our triangle. We also know that the tunnel is semicircular. In other words, the distance between the center of the lane and the edge of the tunnel is the radius of the tunnel which is half the diameter or 8.5 feet. This becomes the hypotenuse of the right triangle. We are interested in finding the height of the tunnel which happens to be the third side in our right triangle. As with any other right triangle, once we know two sides, we know the third. Using Pythagorean's theorem to solve for the height which we will call y, y² = 8.5² - 7.5². Solving for y, we see that y=4. Thus, the height of the tunnel is 4 feet at the edge of the lane.

For the third question, we are told we have an ellipse and are given some lengths. To model this ellipse, we should transform the information we are given to create the standard form ellipse equation. This equation is (x-h)²/a² + (y-k)²/b² = 1, where (x,k) is the center of the ellipse and a and b are the half the lengths of the major (horizontal in this case) and minor (vertical in this case) axes respectively. We can assume the ellipse is centered at (0,0) since we are not given any information that suggests otherwise. Thus, h and k become 0. We also know a and b. a is half the length of the tunnel, 18. b is the height of the tunnel, 15. Note that the height of the tunnel is half the height of the ellipse we are modeling. Following this, the ellipse equation becomes x²/(18²) + y²/(15²) = 1. Now that we know the ellipse, we can figure out any point on the tunnel. Now, let us think about the truck. A truck passing through the center of the tunnel has more headroom than a truck on the sides of the tunnel. Knowing that the truck passes through the center and is 12 feet wide, we know the shortest part of the tunnel is at the edge of the truck, or at +/- 6 feet. Now, we just need to find the height of the tunnel when x is 6 feet. Solving for y in the ellipse equation, we see that y = 14.14 feet. Thus, theoretically a truck can be 14.14 feet tall and still pass through the tunnel. Of course, there are regulations that prevent such close scrapes, but the problem does not ask us to consider those.