Given the equation y=ax
^{2} 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 20111113 12:47:43 