Asked by Andrew
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!
I am soooooo lost!
Answers
Answered by
Steve
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.
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.
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