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Angle of view - Parabola (Posted on 2011-11-13) Difficulty: 3 of 5
Given the equation y=ax2 find the set of all points from which the angle of view* of this parabola is a right angle. What is the significance of this set of points?

* i.e. displaying a right angle between the two tangents.

No Solution Yet Submitted by Jer    
Rating: 3.0000 (1 votes)

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Solution partial solution Comment 2 of 2 |

In the parabola y=x^2:

y'=2x

when

x=a, y=a^2, y'=2a
tangent line is y=2a(x-a)+a^2 = 2ax - a^2

To get y'= -1/(2a),
x=-1/(4a), y = 1/(16a^2)
tangent line is y=(-1/(2a))(x+1/(4a)) + 1/(16a^2)

When the tangents meet:

2ax - a^2 = -x/(2a) - 1/(16a^2)

x = (a^2 - 1/(16a^2))/(2a+1/(2a))
y = 2ax - a^2


Plotting this:

DEFDBL A-Z
CLS
FOR a = -10 TO 10
 IF a <> 0 THEN
  x = (a * a - 1 / (16 * a * a)) / (2 * a + 1 / (2 * a))
  y = 2 * a * x - a * a
  PRINT USING "###.### ####.########"; x; y
 END IF
NEXT a

 -4.988   -0.25000000
 -4.486   -0.25000000
 -3.984   -0.25000000
 -3.482   -0.25000000
 -2.979   -0.25000000
 -2.475   -0.25000000
 -1.969   -0.25000000
 -1.458   -0.25000000
 -0.938   -0.25000000
 -0.375   -0.25000000
  0.375   -0.25000000
  0.938   -0.25000000
  1.458   -0.25000000
  1.969   -0.25000000
  2.475   -0.25000000
  2.979   -0.25000000
  3.482   -0.25000000
  3.984   -0.25000000
  4.486   -0.25000000
  4.988   -0.25000000
 


 So the locus seems to be the line y = -.25 for this parabola.  Unfortunately I didn't incorporate a coefficient for x^2 in the equation, but it would seem that the constant value for y would change in some proportion to the a of the puzzle. (The a of my partial solution is just differing values for x.)  I'll leave it for others to expand on this. My hunch is it's y=-a/4.


  Posted by Charlie on 2011-11-13 13:21:01
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