in the backyard there are two trees located at grid points A(-2,3) and B(4,-6)
a) The family dog is walking through the backyard so that it is at all times twice as far from A as it is from B. Find the equation of the locus of the dog. Draw a graph that shows the two trees, the path of the dog and the relationship defining the locus. Then write a geometric description of the path od the dog relative to the two trees.
b) The family cat is also walking in the backyard. The line segment between the cat and the two trees are always perpendicular. Find the equation of the locus od the cat. Draw a graph that shows the path of the cat. Then write a geometric description of the path of the cat relative to the two trees.
Hey guys, i was wondering if anyone can help me get started with these two questions. Usually, i i don't have that much problems with this chapter, but when dealing with word questions, i get lost. Thanks for your help. Alley
let the dog's position be P(x,y)
Translate the English into math :
"..it is at all times twice as far from A as it is from B"
AP = 2BP
√[(x+2)^2 + (y-3)^2] = 2√[(x-4)^2 + y-6)^2]
square both sides, while expanding at the same time :
x^2 + 4x + 4 + y^2 - 6y + 9 = 4(x^2 - 8x + 16 + y^2 - 12y + 36)
reduces to x^2 - 12x + y^2 - 14y = 65
Do you recognize this to be the equation of a circle?
Can you find the centre and the radius anf draw the sketch or graph from there?
For the second question, if we again let P(x,y) be any position of the cat
isn't the slope of AP always the negative reciprocal of the slope of BP?
Let me know what you got as an equation.
AP = 2BP
�ã[(x+2)^2 + (y-3)^2] = 2�ã[(x-4)^2 + y-6)^2]
maybe it's just me, but shouldn't be
�ã[(x-2)^2 + (y+3)^2] = 2�ã[(x+4)^2 + y-6)^2]
Just saw your reply now.
I did the question on a piece of paper before typing in the solution and
inadvertently wrote down the point B(4,-6) as (4,6)
so the equation I had written as
√[(x+2)^2 + (y-3)^2] = 2√[(x-4)^2 + y-6)^2]
should have been
√[(x+2)^2 + (y-3)^2] = 2√[(x-4)^2 + y+6)^2]
make the necessary changes in my solution, the only change would be in the y term.
Sure, I can help you get started with these two questions!
a) Let's start by finding the midpoint of the line segment AB, which will give us the coordinates of the point equidistant from A and B. The midpoint can be found by averaging the x-coordinates and the y-coordinates separately:
Midpoint coordinates:
x-coordinate: (A's x-coordinate + B's x-coordinate) / 2 = (-2 + 4) / 2 = 1
y-coordinate: (A's y-coordinate + B's y-coordinate) / 2 = (3 + (-6)) / 2 = -1.5
So, the midpoint is located at (1, -1.5).
Now, let's define a point P on the locus of the dog, such that the distance between P and A is twice the distance between P and B. The distance formula between two points (x1, y1) and (x2, y2) is:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Using this formula, we can write the equation for the locus of the dog. Let P be the point (x, y):
Distance AP = 2 * Distance BP
√((x - (-2))^2 + (y - 3)^2) = 2 * √((x - 4)^2 + (y - (-6))^2)
Squaring both sides of the equation and expanding, we get:
(x + 2)^2 + (y - 3)^2 = 4 * ((x - 4)^2 + (y + 6)^2)
Expanding and simplifying further, we arrive at the equation of the locus of the dog:
5x^2 - 38x + 69y - 280 = 0
To graph the locus, plot the points A(-2,3) and B(4,-6), draw the path defined by the equation 5x^2 - 38x + 69y - 280 = 0, and include the two trees.
Geometric description of the path of the dog relative to the two trees: The path of the dog is an ellipse with the two trees as the foci. The dog's distance from one tree is twice its distance from the other tree.
b) To find the locus of the cat, we need to consider that the line segment between the cat and each of the two trees is perpendicular. Let C(x, y) be the point representing the cat's position.
We will use the concept of slopes to find the equation of the locus of the cat. The slope of the line segment between C and A should be negative reciprocal to the slope between C and B for the lines to be perpendicular.
The slope of the line segment between C and A:
Slope CA = (y - 3) / (x - (-2))
The slope of the line segment between C and B:
Slope CB = (y - (-6)) / (x - 4)
Now, we can set up the equation for the perpendicular slopes:
(Slope CA) * (Slope CB) = -1
[(y - 3) / (x + 2)] * [(y + 6) / (x - 4)] = -1
Cross multiplying and simplifying, we get:
(y - 3)(y + 6) = -(x + 2)(x - 4)
Expanding and multiplying further, we arrive at the equation of the locus of the cat:
y^2 + 3y - 18 = -x^2 - 2x + 8
x^2 + y^2 + 2x + 3y - 26 = 0
To graph the locus, plot the points A(-2,3) and B(4,-6), and draw the path defined by the equation x^2 + y^2 + 2x + 3y - 26 = 0.
Geometric description of the path of the cat relative to the two trees: The path of the cat is a circle with the two trees as its diametrically opposite points. The cat's position is always equidistant from the two trees.