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.
In the parabola y=x^2:
y'=2x
when
x=a, y=a^2, y'=2a
tangent line is y=2a(xa)+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 AZ
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 20111113 13:21:01 