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.
https://www.desmos.com/calculator/x85p6kgeet
I used a bit of calculus here. The function f(x)=1/x has derivative f'(x)=-1/x^2. Thus the perpendicular at (x,1/x) has slope x^2.
For a given (k,1/k) on the hyperbola, the perpendicular line has equation y=k^2x+(1-k^4)/k
Any point on the locus must also be on the perpendicular bisector of the point joining (0,0) and (k,1/k). The line has equation y=-k^2x+(1+k^4)/(2k)
The intersection of these lines is ((3k^4-1)/(4k^3)), (3-k^4)/(4k))
The parametric equations are
x(t)= (3t^4-1)/(4y^3)
y(t)= (3-t^4)/(4k)
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Posted by Jer
on 2024-07-29 09:47:17 |
part b)
