Find the equation of the graph of all points such that the difference of their distances from (10,0) and (2,0) is always 1.
hyperbola centered at x =12/2 = 6
hits x axis at (5.5 , 0) and (6.5 , 0)
(x-6)^2/a^2 - y^2/b^2 = 1
2 a = 1
a = 1/2
a^2 = 1/4
4 (x-6)^2 - y^2/b^2 = 1
center to either focus = 4 =sqrt(a^2+b^2)
so
16 = 1/4 + b^2
b^2 = 16 - 1/4 = 63/4
4(x-6)^2 - 4y^2/63 = 1
(x-6)^2 - y^2/63 = 1/4
To find the equation of the graph of all points such that the difference of their distances from (10,0) and (2,0) is always 1, we can follow these steps:
Step 1: Understand the problem:
We need to find all the points that satisfy the condition of having a difference of 1 between their distances from (10,0) and (2,0).
Step 2: Use the distance formula:
The distance between two points (x₁, y₁) and (x₂, y₂) in a coordinate plane is given by the distance formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
Using this formula, we can calculate the distance of any point from (10,0) and (2,0).
Step 3: Set up the equation:
To find the equation of the graph, we need to set up an equation that represents the given condition. Let's assume the coordinates of the point we want to find are (x, y).
According to the given condition, the difference between the distance of (x, y) from (10,0) and (2,0) is always 1. So mathematically, we can write:
|dist((x, y), (10,0)) - dist((x, y), (2,0))| = 1
Step 4: Simplify the equation:
Let's calculate the individual distances using the distance formula:
dist((x, y), (10,0)) = √((10 - x)² + (0 - y)²)
dist((x, y), (2,0)) = √((2 - x)² + (0 - y)²)
Replacing these values in the equation, we get:
|√((10 - x)² + y²) - √((2 - x)² + y²)| = 1
Step 5: Square both sides of the equation:
To eliminate the absolute value, we square both sides:
(√((10 - x)² + y²) - √((2 - x)² + y²))² = 1²
((10 - x)² + y²) - 2√((10 - x)² + y²)((2 - x)² + y²) + ((2 - x)² + y²) = 1
Step 6: Simplify the equation:
Expand and simplify the equation to get rid of the square roots:
(10 - x)² + y² - 2√((10 - x)² + y²)((2 - x)² + y²) + (2 - x)² + y² = 1
(10 - x)² + y² - 2√(((10 - x)² + y²)((2 - x)² + y²)) + (2 - x)² + y² - 1 = 0
This equation represents the graph of all points such that the difference of their distances from (10,0) and (2,0) is always 1.