GEOMETRY question on ellipses and hyperbolas (math wizs are needed)?

Question

Let L be a line in the plane, F by a point in the plane outside of line L. Let X be any point in the same plane, let dL be the distance from X to L, let dF be the distance from X to F.
dF/dL= e
WHERE e is a constant positive number.
Obviously e=1 if it was a parabola.
But 01 is the locus of a hyperbola. So we have the uniform definition of all three conics, depending only on the value of the eccentricity e.

PROVE IT! Starting from the condition (*), express it in coordinated, and by a series of algebraic calculations, to get our standard equation of an ellipse and hyperbola (may be shifted) depending on value of e.

by definition

i mentioned eccentricity and the locus of points of an ellipse and a hyperbola

KK I JUST ADDED THE WHOLE QUESTION AND SCANNED IT TO THE INTERNET, SO EVERYTHING IS CLEAR

http://x60.xanga.com/dccd2a0713c31116227432/b83278373.jpg

Answer 1

let S be the focus, e be the eccentricity and L=0 be the directrix.Let P be a point on the ellipse.Let M,Z be the projections of P and S on the directrix L=0.Let N be the projection of P on SZ.Let A and B be the two ends of major axis of an ellipse and C be its midpoint(centre of the ellipse).Let AB=2a.The points A and B lie on the ellipse.
The locus of a point which moves such that its distance from a fixed point and a fixed line bears a constant ratio is an ellipse(constant ratio is e and 0 hence SA/AZ = e and SB/BZ =e
therefore SA = e AZ and SB = e BZ
SA + SB = e AZ + e BZ implies
AB = e(AZ+BZ)
2a = e(CZ-CA+BA+CZ)
2a =e(2CZ) [since CA=BC]
CZ = a/e
Also SB - SA =eBZ-eAZ
BC+CS-(CA-CS) = e(BZ - AZ)
2CS = eAB
CS = ae
Take CS as the principal axis of the ellipse and CR perpendicular to CS as y-axis.Then
S =(ae,0) and the ellipse is in the standard form.Let P(x1,y1)
PM = NZ =CZ-CN =a/e -x1
P lies on the ellipse i.e,
PS/PM =e
(PS)^2 = e^2 (PM)^2
implies
(x1-ae)^2 +(y1-0)^2 = e^2 [a/e -x1)^2]
(x1-ae)^2 + y1^2 =(a-x1e)^2
x1^2 +a^2e^2-2x1ae+y1^2
=x1^2e^2 +a^2e^2-2x1ae
(1-e^2)(x1^2)+y1^2 =(1-e^2)a^2
x1^2/a^2 + y1^2/(1-e^2)a^2 = 1
Let b^2 = a^2(1-e^2)>0
hence locus of P is
x^2/a^2 + y^2/b^2 =1
therefore the equation of the ellipse is
x^2/a^2 + y^2/b^2 =1


the let S be the focus, e be the eccentricity and L=0 be the directrix.Let P be a point on the hyperbola. Let N be the projection of P on SZ.Let A and B be the two ends of major axis of an ellipse and C be its midpoint(centre of the ellipse).Let AB=2a.The points A and B lie on the hyperbola.
For a hyperbola e>1,
hence SA/AZ = e and SB/BZ =e
therefore SA = e AZ and SB = e BZ
SA + SB = e AZ + e BZ implies
CS-SA+CS+BC =e(AZ + BZ)
2CS = e AB since CA = CB
2CS = e 2a
CS=ae
SB-SA = eBZ-eAZ implies
AB = e(BZ-AZ)
2a = e[CB+CZ-(CA-CZ)]
i.e,2a = e(2CZ)
since CA = CB
CZ = a/e
Take CS as the principal axis of the ellipse and CR perpendicular to CS as y-axis.Then
S =(ae,0) and the hyperbola is in the standard form.Let P(x1,y1).
PM = NZ=CN-CZ = x1 - a/e
P lies on the hyperbola
PS/PM =e
(PS)^2 = e^2 (PM)^2
implies
(x1-ae)^2 +(y1-0)^2 = e^2 [x1-a/e)^2]
(x1-ae)^2 +(y1)^2 =(x1e-a)^2
x1^2 +a^2e^2-2x1ae+y1^2
= x1^2(e^2) +a^2e^2-2x1ae
x1^2(e^2-1) - y1^2 = a^2 (e^2-1)
x1^2/a^2 - y1^2/a^2 (e^2-1) = 1
let b^2 = a^2 (e^2-1)>0
locus of P is x^2/a^2 - y^2/b^2 =1
hence standard form of a hyperbola is
x^2/a^2 - y^2/b^2 =1

Answer 2

Take a system of coordinates in which
L has equation x= a^2/c and F(c,0) (a is still undefined and just used to relate the positions of F and L
The locus is defined by
[(x-c)^2 +y^2]/(x-a^2/c)^2 = (c/a)^2 so
[(x-c)^2+y^2]= (c/a)^2*(x-a^2/c)^2
Making only operations you´ll get the known
equations

Answer 3

uh .... why is it so that if e=1 then "it" is a parabola

I draw the figure and you only mention points and lines , no ellipses

something is wrong / missing in your question