Dimensions of an ellipse region a dog wanders

[aisha]

This is a conic application question worth 8 marks

You don't want your dog to run away so you place 2 steaks in the groud 4 metres apart and attach the end of a 8 metre long rope to each side of the stakes then you attach the dogs collar to a ring which slides along the rope.
Describe the region in which the dog may roam, and include the dimensions fo the region in your answer.

The dimension part of this question is giving me trouble
I know the region will look like a horizontal ellipse
with center (0,0) and steaks (foci) at points (-2,0) and (2,0)

if there was a digram the length from center (0,0) to the foci =c
and the distance from the dog to the foci =a and b is the line perpendicular to the x-axis.

im not sure but I think a=6 and c=2 i used pythagorean theory to find b and got sqrt(32)

now i think the dimensions of the ellipse will be the major axis by the minor axis so 2a by 2b 2(6)=major axis of 12 and 2(sqrt32)=minor axis 11.31

please help me out, thanks

 

[dextercioby]

Nope,the minor axis should come down to 2\sqrt{3}.Can u see why? HINT:There's an equilateral triangle there...

Daniel.

 

[HallsofIvy]

I do hope you are using stakes. The dog would just eat the steaks!

Imagine the dog at the "end of his rope" along the line formed by stakes. measuring from the farther stake, the rope extends between the stakes, a distance of 4 metres, then to the dog (call that "a") then back to the dog, again a distance "a". That's a total of 4+ 2a and must equal the total length of the rope, 8 m. That is, 4+ 2a= 8 so
2a= 4 and a= 2. The dog can go, in line with the stakes, 2 meters beyond the stakes. Since that is true in both directions, the two extreme parts have distance 2+ 4+ 2= 8 m between them- the axis is 8 m long and the "half-axis" is 4 m.

Now imagine the dog going as far as he can perpendicularly to the stakes. If he is exactly between the stakes, by symmetry the rope is equal on both sides of the dog- 4 m on each side. The dog and the two stakes form an isosceles triangle. Dropping a perpendicular exactly between the stakes, we have a right triangle with one leg of length 2 m (half the distance between the stakes) and hypotenus of length 4 (half the length of the rope. By the Pythagorean theorem, the other leg (another half-axis) has length \sqrt{4^2- 2^2}= \sqrt{16- 4}= \sqrt{12}= 2\sqrt{3}.

Okay, choose the x-axis to be the line of the stakes, the y-axis perpendicular to that and (0,0) to be the middle point between the two stakes. You have an ellipse with center at (0,0), x semi-axis (the major semi-axis) of length 4 and y semi-axis (the minor semi-axis) of length 2\sqrt{3}. What is the equation of that ellipse?

 

[aisha]

I think i understand on the x-axis the points of the vertices are (8,0) and (-8,0) I was wondering how do you know the hypotenuse of the right angle triangle is 4m? a=4m is the hypotenuse also a? My foci should be at (-2,0) and (2,0) right? I got the equation of the ellipse to be \frac {x^2} {16} + \frac {y^2} {12} =1

and i think to get the dimensions I need to find the major and minor axis
major=2a=8 and
minor=2b=2(2sqrt3)=6.93m

therefore the dimensions of this ellipse the region in which the dog roams is 8m by 6.93m? Is this how I am supposed to write the dimensions?

 

[aisha]

According to Halls of Ivy's good solution are my dimensions in the last post correct? For the dimensions do i have to write a number by a number or just the equation of the ellipse?

 

[dextercioby]

"Describe the region in which the dog may roam, and include the dimensions fo the region in your answer."

The equation and the parameters should do it:minor & major semiaxis.


Daniel

 

[HallsofIvy]

I was wondering how do you know the hypotenuse of the right angle triangle is 4m?

.
The two hypotenuses, from one focus to the point (0,y) back to the other focus are formed by the rope. You told us the rope the rope was 8 m long so each half of it is 4 m long.

 

[aisha]

So what do you think will I get the full 8 marks? Is the solution correct?