For each point in a set of points, its distance from (3,4) is four times it's distance from (-5,2)
a. Find the equation
b. tell which conic section the graph will be.
let d1 = distance of (x,y) from (3,4)
d1^2 = (x-3)^2 + (y-4)^2
let d2 = distance from (-5,2)
d2^2 = (x+5)^2 + (y-2)^2
since d1 = 4*d2
d1^2 = 16*d2^2
(x-3)^2 + (y-4)^2 = 16[(x+5)^2 + (y-2)^2]
expand and collect terms to arrive at
(x+83/15)^2 + (y-28/15)^2 = 7234/15
Looks like a circle
To find the equation of the set of points that satisfy the given conditions, we can start by assuming a generic point (x, y) in the set.
According to the given conditions, the distance between the point (x, y) and (3, 4) is four times the distance between the point (x, y) and (-5, 2).
To calculate the distance between two points (x1, y1) and (x2, y2), we can use the distance formula:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
So, based on the given conditions, we can write the equation for the distance as follows:
sqrt((x - 3)^2 + (y - 4)^2) = 4 * sqrt((x + 5)^2 + (y - 2)^2)
To eliminate the square roots, we square both sides of the equation:
((x - 3)^2 + (y - 4)^2) = (4 * sqrt((x + 5)^2 + (y - 2)^2))^2
Simplifying further, we expand the squares:
(x - 3)^2 + (y - 4)^2 = 16 * ((x + 5)^2 + (y - 2)^2)
Expanding both sides of the equation:
x^2 - 6x + 9 + y^2 - 8y + 16 = 16 * (x^2 + 10x + 25 + y^2 - 4y + 4)
Simplifying:
x^2 - 6x + y^2 - 8y + 25 = 16x^2 + 160x + 400 + 16y^2 - 64y + 64
Bringing all the terms to one side:
15x^2 + 166x + 15y^2 + 56y - 315 = 0
Therefore, the equation of the set of points that satisfy the given conditions is:
15x^2 + 166x + 15y^2 + 56y - 315 = 0
To determine which conic section the graph will be, we can look at the coefficients of x^2 and y^2. In this case, both coefficients are positive, suggesting that the graph represents an ellipse.