The shape of a park can be modeled by a circle with the equation x^2 + y^2 = 1600. A stretch of highway near the park is modeled by the equation y = 1/40(x - 40)^2. At what points does a car on the highway enter or exit the park?
I am soooooo lost!
If we try to solve this algebraically, we run into trouble.
x^2 + y^2 = 1600
x^2 + (1/40 (x-40))^2 = 1600
x^2 + 1/1600 (x-40)^4 = 1600
(x-40)^4 + 1600x^2 = 1600^2
Now, that's a hard one. However, if you examine the two equations, you see that
the first is a circle of radius 40, which intersects the axes at (40,0) and (0,40)
the second is a parabola with vertex at (40,0) and y-intercept at (0,40)
That make sit easy.
I can help explain how to find the points where the car enters or exits the park.
To find the intersection points, we need to solve the system of equations formed by the circle and the highway equation:
1) x^2 + y^2 = 1600 (Equation of the park)
2) y = 1/40(x - 40)^2 (Equation of the highway)
First, substitute the value of y from the second equation into the first equation:
x^2 + (1/40(x - 40)^2)^2 = 1600
Now we can simplify the equation by expanding and collecting like terms:
x^2 + (1/40)^2(x - 40)^4 = 1600
Next, multiply both sides by 40^2 to remove the fraction:
40^2x^2 + (x - 40)^4 = 1600 * 40^2
Simplify further and expand (x - 40)^4 term:
40^2x^2 + (x^4 - 4x^3 * 40 + 6x^2 * 40^2 - 4x * 40^3 + 40^4) = 1600 * 40^2
Now combine like terms and rearrange the equation:
x^4 - 4x^3 * 40 + 6x^2 * 40^2 - 4x * 40^3 + 40^4 + 40^2x^2 - 1600 * 40^2 = 0
After simplifying and collecting like terms:
x^4 - 4x^3 * 40 + 7x^2 * 40^2 - 4x * 40^3 + 40^4 - 1600 * 40^2 = 0
This is a quartic equation that can be solved using numerical methods or software. Once you have solved for x, you can substitute it back into either of the two equations to find the corresponding y value. These points (x, y) represent the locations where the car enters or exits the park, depending on which side of the highway the points lie.