Hint: A locus of points equidistant from a point and a line is a parabola. A quadratic equation models a parabola.
P(x;y). Is a point which is equidistant to both point D(2;-5) and straight line y=1
Determine the equation of locuss of P(x;y)
I drew horizontal y=1 and plot D(2;-5). Am not sure how to draw locus of points which satisfies given conditions?
Thanks in advance!
The distance between two points is d^2 = x^2 + y^2 where x is the x-distance and y is the y-distance.
For a circle this is (x-a)^2 + (y-b)^2 = r^2 and for the line you have y = 0 or y = 2 (since in euclidean space, you have parallel lines for constant distance).
So now you have three equations and you must solve for these.
@ Chiro I think u r complicating the problem.I did try using definitions of loci,after having (1) equation of circle => (x-2)^2 +(y+5)^2 = 29 (2) y=0 and (3) y=2. How do I work the equation because the equation must in the form of y = ax^2 +bx+c. Its more of manipulating than going with definitions. If I get my sketch right I can equate the distances from the line y=1 to P(x;y) and from P(x;y) to D(2;-5)
Lot many suggestions have been given i would like to just add that we have an expression for finding distance of a point ( p,q) from a line ax+by+c=0 given by
[ |ap+bq+c|]/[ sqrt ( a^2 +b^2)] Use this and the other expression given for the distance between two points and you will get the locus.
This another attempt of mine after many suggestions. I set a third point along the line y=1 I call this point Q(x;1) then distance from PQ =PD i.e PQ^2 =PD^2
d^2 =(x2-x1)^2 +(y2-y1)^2 substituting we have
(x-x)^2 + (y-1)^2 = (2-x)^2 + (-5-y)^2
y^2 -2y +1=x^2 -4x +4 +y^2 +10y +25
12y = -x^2 +4x -28
Please verify this. thanks in advance