- #1
physicsgal
- 164
- 0
here's the problem:
in a backyard, there are two trees located at grid points A(-2, 3) and B(4,-6)
a) a dog walks 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.
so PA = 2(PB)
(x+2)^2 + (y-3)^2 = 4((x-4)^2 + (y+6)^2)
i asked a teacher and they said it's a ellipse and the equation is
5x^2(2) - 28x + 2y^2 + 6y + 113
but that doesn't make sense because the course hasnt dealt with ellipses yet, just lines and circles. does it have to be an ellipse?
b) the family cat is also walking in the backyard. the line segments between the cat and the two trees are always perpendicular. find the equation for the locus of the cat.
so this means it path is equidistant from AB?
so PA = PB
(x+2)^2 + (y-3)^2 = (x-4)^2 + (y+6)^2
with my math i end up with y = 2.17 - 0.67x.. but this doesn't look right. there must be something wrong.
any help is appreciated!
~Amy
in a backyard, there are two trees located at grid points A(-2, 3) and B(4,-6)
a) a dog walks 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.
so PA = 2(PB)
(x+2)^2 + (y-3)^2 = 4((x-4)^2 + (y+6)^2)
i asked a teacher and they said it's a ellipse and the equation is
5x^2(2) - 28x + 2y^2 + 6y + 113
but that doesn't make sense because the course hasnt dealt with ellipses yet, just lines and circles. does it have to be an ellipse?
b) the family cat is also walking in the backyard. the line segments between the cat and the two trees are always perpendicular. find the equation for the locus of the cat.
so this means it path is equidistant from AB?
so PA = PB
(x+2)^2 + (y-3)^2 = (x-4)^2 + (y+6)^2
with my math i end up with y = 2.17 - 0.67x.. but this doesn't look right. there must be something wrong.
any help is appreciated!
~Amy