conics (1 Viewer)

[gronkboyslim]

P(a cos @ , bsin @) and Q[acos (-@), b sin (-@)] are extremities of the latus rectum x = a/e of the ellipse x^2/b^2 +y^2/b^2 = 1

1) show that PQ has length 2(b^2/a)

help

 

[Mordenkainen]

Originally posted by gronkboyslim
P(a cos @ , bsin @) and Q[acos (-@), b sin (-@)] are extremities of the latus rectum x = a/e of the ellipse x^2/b^2 +y^2/b^2 = 1

1) show that PQ has length 2(b^2/a)

help

i haven't touched 4u maths in a LONG time, so i could be wrong here. there seem to be 2 flaws
a) x^2/b^2 + y^2/b^2 = 1 is a circle, not an ellipse (unless a circle is an ellipse )
b) isn't x = a/e the directrix eqn? if so, it wouldn't be a latus rectum of the "ellipse" (since a latus rectum is a perpendicular chord or something), and therefore wouldn't touch the circle at all

 

[spice girl]

It's prolly:
latuus rectum x = ae, of the general ellipse x^2/a^2 + y^2/b^2 = 1

Sub in x = ae: you have (ae)^2/a^2 + y^2/b^2 = 1
y^2/b^2 = 1 - e^2
y^2 = b^2(1-e^2)

for ellipse, we have b^2 = a^2(1 - e^2)
(1 - e^2) = b^2/a^2

y^2 = b^2(b^2/a^2)
so y = (+-)(b^2/a)

the distance PQ = |y1 - y2| where y1,y2 are the two solutions to the above quadratic in y.

PQ = b^2/a - (-b^2/a) = 2b^2/a

 

[gronkboyslim]

thanx