A landscaper wants to plan a walkway that passes between a tree and the border of the lawn. Using these as the focus and directrix, how can the landscaper plan a parabolic path that will be equidistant from the tree and the border at all times? Describe your method in full sentences.

Question

anonymous · Answer

@campbell_st

campbell_st · Answer

well the tree is the F the focus and the path is the directrix y = m .... you can create an equation ising the distance formula 

well have a point P(x, y) on the parabola 

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as P moves along the parabola the distance FP is always equal to PB

B has the same x value as P and the  y value is always fixed by the directrix. 

So if $$F(x_{1}, y_{1},..and.. B(x, m)$$

becuase the point P is always equidistant

FP = PB  

$$\sqrt{ x - x_{1})^2 + (y - y_{1})^2 } = \sqrt{(x - x)^2 + (y - m)^2}$$

if you square both sides of the equation and then distribute and simplify you'll get the equation of the parabola. 

use the distance formula for FP and BP