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.

Let (x_0,y_0) be the intersection of the 
tangents and (x_1,y_1) and (x_2,y_2) the 
points of tangency.

Parabola y = a*x^2 with y' = 2*a*x.

For point (x_2,y_2) we have

  (y_2 - y_0)/(x_2 - x_0) = 2*a*x_2

    or

  (a*x_2^2 - y_0)/(x_2 - x_0) = 2*a*x_2

    or

  -y_0 = a*x_2^2 - 2*a*x_0*x_2        (1)

Similarly for point (x_1,y_1) we have
  
  -y_0 = a*x_1^2 - 2*a*x_0*x_1        (2)

Subtracting (2) from (1) gives

  0 = a*(x_2 - x_1)*(x_2 + x_1 - 2*x_0)

Since a != 0 and x_2 != x_1,

  2*x_0 = x_1 + x_2                   (3)

Plugging (3) into either (1) or (2) gives

  y_0 = a*x_1*x_2                     (4)

Since the tangents are orthogonal,

  (2*a*x_1)*(2*a*x_2) = -1

    or

  4*a^2*x_1*x_2 = -1                  (5)

Combining (4) and (5) gives

  y_0 = -1/(4*a)

Combining (3) and (5) gives

  4*a^2*x_2^2 - 8*a^2*x_0*x_2 - 1 = 0

Solving for x_2 gets

  x_2 = x_0 + sqrt(4*a^2*x_0^2 + 1)/(2*a)

Therefore, for any real number x_0 there
exists a real number x_2 such that point
(x_0,-1/(4*a) is in the locus.

Thus, the line

  y = -1/(4*a)                        (6)

is the locus of points and it is also
the directrix of the parabola.

 

Posted by Bractals on 2011-11-13 12:47:43