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(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.
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Posted by Charlie
on 2011-11-13 13:21:01 |
partial solution
