The points equidistant from (0, 0) and (1, 1) form the perpendicular bisector of the segment joining the given points, which has the equation x + y = 1.
The distance from a point to a line or curve is defined as the shortest distance.
Call the equidistant point (x, y) and the closest point on the curve (a, b).
a) The set (locus) of points equidistant from (0, 0) and the line x + y = 1 forms a familiar shape.
Name the shape, and find its equation.(as a relation between x and y or as parametric equations of a)
b) Determine the locus of points equidistant from the origin and the hyperbola x y = 1, x > 0. (as a relation between x and y or as param. eqns of a)
c) Find the locus of points equidistant from the origin and the circle through (1, 1) that's tangent to both axes. (in {x, y} or a form)
Show your steps and reasoning.
Edit to share graph: https://www.desmos.com/calculator/rlzjluleks
Since we know this is a parabola, it makes sense to find points on lines parallel to the axis of symmetry (y=x), since each line will have one point.
points on y=x+k have coordinates (x,x+k)
the line will be perpendicular to x+y=1 at (1/2-k/2,1/2+k/2)
distance from origin = sqrt(x^2+(x+k)^2)
distance from the line = sqrt((x-1/2+k/2)^+(x+k-1/2-k/2)^2)
Setting these equal gives
x=-k^2/4-k/2-1/4
y=-k^2/4+k/2+1/4
The parametric equations are
x(t)=-t^2/4-t/2-1/4
y(t)=-t^2/4+t/2+1/4
(Note: setting the distances equal is convenient since squaring gets rid of the square roots. The same won't happen on parts b) and c))
Edited on July 29, 2024, 10:50 am
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Posted by Jer
on 2024-07-29 08:57:31 |
part a)
