 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Angle of view - Parabola (Posted on 2011-11-13) 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) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 2
`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 thereexists 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 alsothe directrix of the parabola.`

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