More conics application

[peep1963]

HHHeeellllpppp!!!!!!!!!!!!!!

My math teacher gave us a 25 Conics(ugh!) application problem sheet as a Christmas gift :evil: and I can't figure out these! Math gods, please explain so a poor mortal like myself can understand.


1. A satellite orbits around the moon in an elliptical path. Assuming the moon to be at the center of a rectangular cooridinate system, find the equation of the elliptical path of a satellite whose x-intercepts are +/- 82,000 km and whose y-intercepts are +/- 71,000 km.



2. Suppose you are Cheif Mathmatecian for Ornery and Sly Construction Company. Your company has a contract to build a football stadium in the form of two concntric ellipses, with the fielsd inside the inner ellipse, and the seats between two elipses. The seats are in the intersection of the graphs of X^2 + 4Y^2 is greater than or equal to 100 AND 25X^2 + 36Y^2 is less than or equal to 3600, where each unit on the graph represents 10 meters.

A. Draw a graph of the seating area.

B. The area of an elliptical region is pi(ab), where a and b are the semi-major/minor-axis respectively. The Engineering Dept. estimates that each seat occupies 0.8 square meters. What is the seating capaciy of the stadium?



3. Suppose you have been hired by Palomar Observatory near San Diego. Your assignment is to track incoming meteorites to find out whether or not they will strike the Earth. Since the Earth has a circular cross-section, you decide to set up a Cartesian Coordinate System with its origin at the center of the Earth. The equation of the Earth's surface is X^2 + Y^2 = 40, where x,y, and r, are distances in thousands of kilometers.

A. The first meteorite you observe is moving along the parabola whose equation is 18X- Y^2 = -144. Will this meteorite strike the Earth's surface? If so, where? If not, how do you tell?

B. The second meteorite is coming in from the lower left along one branch of the hyperbola 4X^2 - Y^2 - 80X = -340. Will it strike the Earth's surface? If so, where? If not. how do you tell?

C. To the nearest 100 kilometers, what is the radius of the Earth?

Thanks!

 

[Unco]

1. A satellite orbits around the moon in an elliptical path. Assuming the moon to be at the center of a rectangular coordinate system, find the equation of the elliptical path of a satellite whose x-intercepts are +/- 82,000 km and whose y-intercepts are +/- 71,000 km.

           * + * 71000
       *     |     *
    *        |        * 
  *          |          *
 +-----------+-----------+
  *          |          *|
    *        |        *  82000
       *     |     *
           * + *

The equation for an ellipse centred on the origin is
. . \(\displaystyle \L \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Here, \(\displaystyle \L a = 82000\) and \(\displaystyle \L b = 71000\).

 

[tkhunny]

1. A satellite orbits around the moon in an elliptical path. Assuming the moon to be at the center of a rectangular cooridinate system, find the equation of the elliptical path of a satellite whose x-intercepts are +/- 82,000 km and whose y-intercepts are +/- 71,000 km.

That's pretty much a definition problem, isn't it?

x^2/a^2 + y^2/b^2 = 1

Fill in the blanks.

I am not encouraged that you couldn't get this one. It does not bode well for the rest of them.

 

[galactus]

For #2, they give you :

$\displaystyle x^{2}+4y^{2}\geq100$

$\displaystyle 25x^{2}+36y^{2}\leq3500$

Divide through by the right side of the inequality signs:

$\displaystyle \frac{x^{2}}{100}+\frac{y^{2}}{25}\geq1$

$\displaystyle \frac{x^{2}}{144}+\frac{y^{2}}{100}\leq1$


$\displaystyle \frac{x^{2}}{(10)^{2}}+\frac{y^{2}}{(5)^{2}}\geq1$

$\displaystyle \frac{x^{2}}{(12)^{2}}+\frac{y^{2}}{(10)^{2}}\leq1$

This gives you a and b of the two ellipses. You have the formula for their areas, $\displaystyle {\pi}ab$.
Don't forget about the 'each unit represents 10 meters'.

 

[soroban]

Hello, peep1963!

#3 involves only the intersections of curves . . .

3. Suppose you have been hired by Palomar Observatory near San Diego.
Your assignment is to track incoming meteorites to find out whether or not they will strike the Earth.
Since the Earth has a circular cross-section, you decide to set up a Cartesian coordinate system
with its origin at the center of the Earth.
The equation of the Earth's surface is $\displaystyle \,x^2\,+\,y^2\:=\:40$,
where $\displaystyle x,\,y,$ and $\displaystyle r$ are distances in thousands of kilometers.

A. The first meteorite you observe is moving along the parabola whose equation is $\displaystyle 18x\,-\,y^2\:=\:-144$
Will this meteorite strike the Earth's surface?
If so, where?$\displaystyle \;\;$ If not, how do you tell?

Solve the system: $\displaystyle \:\begin{array}{cc}x^2\,+\,y^2\:=\:40\\18x\,-\,y^2\:=\:-144\end{array}$

The second equation gives us: $\displaystyle \,y^2\:=\:18x\,+\,144$

Substitute into the first equation: $\displaystyle \,x^2\,+\,(18x\,+\,144)\:=\:40$

We have the quadratic: $\displaystyle \,x^2\,+\,18x\,+\,104\:=\:0$

$\displaystyle \;\;\;$The Quadratic Formula gives us: $\displaystyle \,x\:=\:\frac{-18\,\pm\,\sqrt{-92}}{2}$ . . . complex roots.

$\displaystyle \;\;\;$Hence, the two curves do not intersect.


Therefore, the meteorite will not stike the Earth . . . whew!

B. The second meteorite is coming in from the lower left
along one branch of the hyperbola $\displaystyle 4x^2\,-\,y^2\,-\,80x\:=\:-340$
Will it strike the Earth's surface?
If so, where?$\displaystyle \;\;$ If not, how do you tell?

Solve the system: $\displaystyle \,\begin{array}{cc}x^2\,+\,y^2\:=\:40 \\ 4x^2\,-\,y^2\,-\,80x\:=\:-340\end{array}$

The second equation gives us: $\displaystyle \,y^2\:=\:4x^2\,-\,80x\,+\,340$

Substitute into the first equation: $\displaystyle \,x^2\,+\,(4x^2\,-\,80x\,+\,340)\:=\:40$

We have the quadratic: \(\displaystyle \,4x^2\,-\,80x\.+\,300\:=\:0\;\;\Rightarrow\;\;x^2\,-\,16x\,+\,60\:=\:0\)

\(\displaystyle \;\;\;\\)which factors: \(\displaystyle \,(x\,-\,6)(x\,-\,10)\:=\:0\) . . . and has roots: $\displaystyle \,x\,=\,6,\,10$

We must disregard $\displaystyle x\,=\,10$ . . . it's not on the Earth at all!

But the meteorite will strike at $\displaystyle (6,-2).\;$ . . . You've been warned. **

C. To the nearest 100 kilometers, what is the radius of the Earth?

You don't need fancy math for this one . . .

The circle $\displaystyle \,x^2\,+\,y^2\:=\:40\,$ has a radius of $\displaystyle \sqrt{40}\,=\,6.32455532$

This number is in thousands of kilometers.
$\displaystyle \;\;$so the radius of the Earth is: $\displaystyle 6324.55532\text{ km }\approx\:6300\text{ km.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** Really?

I sketched the graphs . . . (No, I don't use a "grapher"!)

The Earth is an origin-centered circle with radius $\displaystyle \sqrt{40}\,\approx\,6.32$
$\displaystyle \;\;\;$It extends to the right as far as $\displaystyle (6.32,\,0)$.


I completed-the-square on the equation of the hyperbola:

$\displaystyle \;\;\;4x^2\,-\,80x\,-\,y^2\:=\:-340$

$\displaystyle \;\;\;4(x^2\,-\,20x)\,-\,y^2\:=\:-340$

$\displaystyle \;\;\;4(x^2\,-\,20x\,+\,100)\,-\,y^2\:=\:-340\,+\,400$

$\displaystyle \;\;\;4(x\,-\,10)^2\,-\,y^2\:=\:60$

$\displaystyle \;\;\;\frac{(x-10)^2}{15}\,-\,\frac{y^2}{60}\:=\:1$

The hyperbola has its center at: $\displaystyle (10,\,0)\:$and its vertices at: $\displaystyle (10\pm\sqrt{15},\,0)$

The meteorite is coming up the lower-left branch of the hyperbola
$\displaystyle \;\;\;$up to its x-intercept: $\displaystyle \,10-\sqrt{15}\,\approx\,6.127$
$\displaystyle \;\;\;$which is inside the Earth's circle.

Darn, there goes my weekend . . .