I need help too on ............ CONICS!

[dummy74]

Hi, I'm new! Uh, I relly don't know how you guys work so I'll just shoot and you guys tell me if I'm doing it right. I should warn you that I am computer illiterate.

Here are my conics questions from homework over the break. They're mostly parabolas and ellipses, both on which I missed the lessons.

A. The US Capital building contains an elliptical room. It is 96ft in length and 46ft in width. Write an equation for the room. Center is at origin and major axis is horizontal. John Quincy Adams discovered that he could overhear conversations being held at the opposing party leader's desk if he stood in a certain spot in the elliptical chamber. Describe the position of the desk and how far away Adams had to stand to overhear.

I think the equation is X^2/96 + Y^2/46 =1. But then what?


B. A lithtripter can be used to break up kidney stones. It uses the properties of ellipses. An electrode sends shock waves out from one focus for an ellipse. The waves are then refelcted off an elliptical surface to the other focus where the kidney stone is located, shattering the stone. suppose length of the major axis is 40 cm and minor axis is 20 cm. How far from the kidney stone should the electrode be places in order to shatter it?

Again, I think the equation is X^2/40 + Y^2/20 =1. Where do I go after that?


C. The cross section of a television antenna dish is a parabola. For one dish, the receiver is located at the focus, 4 ft above the vertex. Find an equation, for cross section of the dish (vertex at the origin). If the dish is 8 ft wide how deep is it?

D. You can make a solar hot dog cooker using foil-lined cardboard shaped as a parobolic trough. If trough is 12 in. wide and 4 in. deep, how far from the bottom should the wire be placed?

E. A spacecraft is in circular orbit (clockwise) 150 km above Earth. Once it obtains the velocity needed to escape Earth's gravity, the spacecraft will follow a parabolic path with focus at the center of the Earth. Supose it obtains its escape velocity above the North Pole. Assume center is at origin and the radius of the Earth is 6400 km. Write an equation for the parabolic pathe of the spacecraft.

F. The towers of the Golden Gate Bridge connecting San Francisco to Marin County are 1280 meters apart and rise 160 meters above the road. The cable between the towers has the shape of a parabola and the cable just touches the sides of the road, midway between the towers. What is the height of the cable 200 meters from a tower?

G. A semi-elliptical (center at the origin) archway over a one-way road has a height of 10 ft and a width of 40 ft. Your truck has a width of 10 ft. and a height of 9 ft. Will your truck clear the openeing of the archway?

I put no, but can someone explain the real answer?

 

[Unco]

G'day.

A. The US Capital building contains an elliptical room. It is 96ft in length and 46ft in width. Write an equation for the room. Center is at origin and major axis is horizontal. John Quincy Adams discovered that he could overhear conversations being held at the opposing party leader's desk if he stood in a certain spot in the elliptical chamber. Describe the position of the desk and how far away Adams had to stand to overhear.

I think the equation is X^2/96 + Y^2/46 =1. But then what?

The general equation of an ellipse centred on the origin
. . . is \(\displaystyle \L \frac{x^2}{a^2} \, + \, \frac{y^2}{b^2} \, = \, 1\)
where \(\displaystyle \L a\) is half the length of the horizontal axis (ie. the length of the semi-major axis); \(\displaystyle \L b\) is half the length of the vertical axis (ie. the length of the semi-minor axis) of the ellipse.

The ellipse we are given has length (the length of the horizontal/major axis) 96 , so \(\displaystyle \L a\) is half that: 48.

Its width (the length of the vertical/minor axis) is 46, so \(\displaystyle \L b\) is half of that: 23.

Its equation is therefore
. . . \(\displaystyle \L \frac{x^2}{48^2} \, + \, \frac{y^2}{23^2} \, = \, 1\)

For the next part of the question, I assume the opposing party leader's desk is located at one of the foci of the ellipse.

A special property of the ellipse is that any ray sent from a focus to the ellipse's surface will reflect back to the other focus.

So in this context, if Mr Adams sits at the other focus, sound waves from the opposition will reflect from the walls of the elliptical chamber to him.

\(\displaystyle \L c\), the distance from the origin to either foci is given by

. . . \(\displaystyle \L a^2 \, = b^2 \, + \, c^2\)

. . . or \(\displaystyle \L c = \sqrt{a^2 \, - b^2}\)

So, here we have

. . . \(\displaystyle \L c = \sqrt{48^2 \, - \, 23^2} \, = \, \sqrt{1775} \, \approx \, 42.1\)

Coordinate-wise, if the opposition is at (42.1, 0), Mr Adams should sit at (-42.1, 0)

 

[galactus]

For F, the Golden Gate problem.

You can use the formula:

\(\displaystyle y=a(x-h)^{2}+k\)

If we center the point where the cable touches the bridge on the origin. we can call that the vertex(0,0).

Since the bridge is 1280' long, it is 640' on either side of the y-axis.

Since a column is 160' high, this gives us a y coordinate. So, we have a point on the parabola with coordinate (640,160).

Using the formula, we first find 'a':

\(\displaystyle 160=a(640-0)^{2}+0\)

\(\displaystyle \frac{160}{640^{2}}=\frac{1}{2560}=a\)

So, the parabola equation is:

\(\displaystyle y=\frac{1}{2560}x^{2}\)

Since we want the height 200 m from a tower, we must subtract from 640. So x=440.

\(\displaystyle \frac{1}{2560}(440)^{2}=\frac{605}{8}=75.625\) meters


For G:

Find an equation for the ellipse:

Since the road is 40 feet wide, it is 20 feet on either side of the y-axis.

The height is 10', therefore, the equation of the ellipse is:

\(\displaystyle \frac{x^{2}}{(20)^{2}}+\frac{y^{2}}{(10)^{2}}=1\)

\(\displaystyle \frac{x^{2}}{400}+\frac{y^{2}}{100}=1\)

Use your x and y values for the truck to see if it can pass through or not.

 

[mmm4444bot]

Uh, I relly don't know how you guys work

Please see the post titled "Read Before Posting" for general guidance on how to ask for help here; in particular, please start a new thread for each exercise.

Cheers ~ Mark :cool:

 

[guitarguy]

For part C of your question.

The cross section of a parabolic antenna will trace out the curve x^2/4f where f is the focus which you are given. If you graph the function with the y-axis as the depth of the antenna for the given value of x.

Keep plugging these problems are tough. Write if you need more help.