sum and difference locus is same parabola?

[gudi]

Point P(0,a) is on y-axis, point M (x,0) is on x-axis, P is in x-y
plane.

If the sum of distances PM and PF equals s, locus of P is

y + sqrt( (y-a)^2 +x^2 ), or,

2 y = s - (x^2 +a^2 -2 a y)/s.

If the difference of distances PM and PF equals s,

then also locus of P, due to squaring of the square root, again is:

2 y = s - (x^2 +a^2 -2 a y)/s.

How can both loci be the same ?

Regards

Narasimham

[gudi]

Sorry, the typo is corrected and y equation made explicit:

If the sum of distances FM and PF equals s, locus of F is

y = (s + a) - x^2 /(s - a)

If the difference of distances FM and PF equals d, then

y = (d + a) - x^2 /(d - a)

How can both loci be parabolas qualitatively ?

Narasimham

[Philippe 92]

Dear Narasimham,

I finally answer on the forum. However it seems there are even fewer
posters here than on geometry.puzzles !

It looks very strange to name F the current point on a parabola
and P its focus ! I restate the problem with usual notations :

Let F the fixed point (0,a) on the y axis
Let P a point (x,y) on the plane and M(x,0) its perpendicular
projection on Ox axis.
To find the loci of P with |PM +/- PF| = given d

First of all, the calculations. The flaw that we have to carefully
avoid is that distance PM is not y but |y|, that is depending on the
sign of y :
if y > 0 : PM = y
if y < 0 : PM = -y

that is we get ||y| +/- sqrt(x^2 + (y - a)^2)| = d
That is the set of the *two* parabolas
(y +/- d)^2 = x^2 + (y - a)^2, or after reduction :

2y(a +/- d) = x^2 + a^2 - d^2, that you may write
2y = x^2/(a - d) + (a + d) (1st parabola)
2y = x^2/(a + d) + (a - d) (2nd parabola)

(you missed the factor 2 in your formulas)

But ! the searched loci are disjoint *parts* of this global locus :
PM + PF = d : PM = |y| < d
|PM - PF| = d : PM = |y| > d
and one or the other parabola depending on the sign of y and of
the sign of PM - PF.

Because the direction of the parabolas depend on the sign of a +/- d,
we finally get the following loci (See attachement) :

1st case, d < a :
================
Locus of PM + PF = d :
2y = x^2/(a - d) + (a + d) with 0 < y < d, or
2y = x^2/(a + d) + (a - d) with -d < y < 0
No such points, the locus doesn't exist.

Locus of PM - PF = d :
2y = x^2/(a - d) + (a + d) with y > d : this full parabola
2y = x^2/(a + d) + (a - d) with y < -d : no such points
The locus is just the parabola 2y = x^2/(a - d) + (a + d)

Locus of PF - PM = d :
2y = x^2/(a - d) + (a + d) with y < -d : no such points
2y = x^2/(a + d) + (a - d) with y > d : this full parabola
The locus is just the parabola 2y = x^2/(a + d) + (a - d)

2nd case, d > a :
================
Locus of PM + PF = d :
2y = x^2/(a - d) + (a + d) with 0 < y < d
2y = x^2/(a + d) + (a - d) with -d < y < 0
The locus is two arcs of two parabolas with |x| < sqrt(d^2 - a^2)

Locus of PM - PF = d :
2y = x^2/(a - d) + (a + d) with y > d : no such points
2y = x^2/(a + d) + (a - d) with y < -d : no such points
The locus doesn't exist

Locus of PF - PM = d :
2y = x^2/(a - d) + (a + d) with y < 0 < d
2y = x^2/(a + d) + (a - d) with y > 0 > - d
The other arcs of the same parabolas, that is |x| > sqrt(d^2 - a^2)

Best Regards.

Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/125-2220165-7353153-663639/2parabolas.gif

[Philippe 92]

Dear all,

The attachment in my previous answer has disapeared.
I upload the attachment again.

Abstract :

Let F the fixed point (0,a) on the y axis
Let P a point (x,y) on the plane and M(x,0) its perpendicular
projection on Ox axis.
To find the loci of P with |PM +/- PF| = given d

The locus is parts of two parabolas, depending on
PM + PF = d, PM - PF = d or PF - PM = d, and also on d < a or d > a

Regards.
Philippe.

Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/125-2220165-7370622-665663/2parabolas.gif