A curve is traced by a point p(x,y) which moves such that its distance from the point a(-1,1) is three times its distance from t
he point b(2,-1). determine the equation of the curve.
1 answer:
Refer to the figure shown below.
The point c, on line segment ab, is in a 3:1 ratio from a.
Therefore,
the x-coordinate of c is -1 +(3/4)*(3) = 5/4 = 1.25.
the y-coordinate of c is 1 - (3/4)*(2) = -1/2 = -0.5.
At point p(x,y), we want d1 = 3d₂, or d₁² = 9d₂².
Therefore
(x + 1)² + (y - 1)² = 9[(x - 2)² + (y + 1)²]
x² + 2x + 1 + y² - 2y + 1 = 9x² - 36x + 36 + 9y² + 18y + 9
8x² - 38x + 8y² + 20y + 43 = 0
x² - 4.75x + y² + 2.5y + 5.375 = 0
(x - 2.375)² - 5.6406 + (y + 1.25)² - 1.5625 + 5.375 = 0
(x - 2.375)² + (y + 1.25)² = 1.828
This is a circle with center at (2.375, -1.25) and radius 1.352.
Answer:
(x - 2.375)² + (y +1.25)² = 1.828
The point c, on line segment ab, is in a 3:1 ratio from a.
Therefore,
the x-coordinate of c is -1 +(3/4)*(3) = 5/4 = 1.25.
the y-coordinate of c is 1 - (3/4)*(2) = -1/2 = -0.5.
At point p(x,y), we want d1 = 3d₂, or d₁² = 9d₂².
Therefore
(x + 1)² + (y - 1)² = 9[(x - 2)² + (y + 1)²]
x² + 2x + 1 + y² - 2y + 1 = 9x² - 36x + 36 + 9y² + 18y + 9
8x² - 38x + 8y² + 20y + 43 = 0
x² - 4.75x + y² + 2.5y + 5.375 = 0
(x - 2.375)² - 5.6406 + (y + 1.25)² - 1.5625 + 5.375 = 0
(x - 2.375)² + (y + 1.25)² = 1.828
This is a circle with center at (2.375, -1.25) and radius 1.352.
Answer:
(x - 2.375)² + (y +1.25)² = 1.828
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Answer:
- a) 3/5·((-2)^n + 4·3^n)
- b) 3·2^n - 5^n
- c) 3·2^n + 4^n
- d) 4 - 3 n
- e) 2 + 3·(-1)^n
- f) (-3)^n·(3 - 2n)
- g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19
Step-by-step explanation:
These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.
If there is a solution of the form , then it will satisfy ...
Rearranging and dividing by , we get the quadratic ...
The quadratic formula tells us values of r that satisfy this are ...
We can call these values of r by the names r₁ and r₂.
Then, for some coefficients p and q, the solution to the recurrence relation is ...
We can find p and q by solving the initial condition equations:
These have the solution ...
_____
Using these formulas on the first recurrence relation, we get ...
a)
__
The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)
_____
For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...
The initial condition equations are now ...
and the solutions for p and q are ...
__
Using these formulas on problem (d), we get ...
d)
__
And for problem (f), we get ...
f)
_____
<em>Comment on problem g</em>
Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.
One angle be x
- other is 2x-6
ATQ