a little help with a math formula please...

[slugg]

I need to know how to find the measure of the latus rectum in an ellipse. The forumula is:

(2b^2) / a = c

where a is the major radius, b the minor, and c the measure of the latus rectum.

What i need to know is how this formula works; how is it derived?

sorry, this is a little off topic, but this seems like the best forum for this question...

 

[Inflamed]

What are you allowed to work from? Which formulas can you take for granted? Or are you supposed to derive the formula by calculus...?



If you work from the formula for a focal point's distance, the formula for the latus rectum can be determined by calculating it as a return value on the ellipse function... my math skills are very rusty, but...


If the distance for the focal point is d=(a^2-b^2)^1/2
And the formula for an ellipse is x^2/a^2 + y^2/b^2 =1

We can reorder the ellipse formula so that y= (b^2 - b^2*x^2/a^2)^1/2

Now, the latus rectum is going to be a value of y, rather twice the height of y where x is the focal point :evil: - the value for y is going to be the same as the height of the latus rectum from the x axis... and the latus rectum will extend across this axis so it's twice that value...


so...

c/2 = y
and we're going to run the function at the point where x=d
x=(a^2 - b^2)^1/2
thus...


c/2= (b^2 - b^2*(a^2 - b^2)/a^2)^1/2

c/2 = (b^2 - b^2*a^2/a^2 + b^4/a^2)^1/2

c/2 = (b^2 - b^2 +b^4/a^2)^1/2

c/2 = (b^4/a^2)^1/2

c/2 =b^2/a

c = 2b^2/a

 

[gorff]

refer to this page for diagrams http://mathworld.wolfram.com/Ellipse.html

refer to 1st figure for variable designations F1, F2, r1, r2, c, a(semi major axis), b(semi minor axis) although they are not labelled on fig2 they are the same.
refer to fig2 and draw a line through F2 parallel to the directrix and the point(upper point) where it intersects the ellipse is the point(P) we will use to derive the semi latus rectum.
reiterating: r1 is distance from F1 to P etc
since by dfn of an ellipse r1 + r2 = 2a
in this case r2 is the semi latus rectum
using the right triangle formed by r1, r2 and 2c -> r1^2 - (2c)^2 = r2^2
or r1^2 - r2^2 = (2c)^2 or (r1 + r2)(r1-r2) = (2c)^2
[if you chose a point on the semi minor axis and did the same you will find that b^2 + c^2 = (2a/2)^2 = a^2
so that c^2 = a^2 - b^2 ]
from (r1 + r2)(r1-r2) = (2c)^2 substitute r1 + r2 = 2a so find 2a(r1-r2) = (2c)^2 = 4c^2 = 4(a^2 - b^2) as in above line.
solve 2a(r1-r2) = 4(a^2 - b^2) for r2 (r2 = r1 - 2(a^2 -b^2)/a )
since r1 = 2a - r2 have r2 = 2a - r2 - 2(a^2 -b^2)/a
simplify to get r2 = a - (a^2 -b^2)/a
simplify further r2 = (a^2 - a^2 +b^2)/a = (b^2)/a = semi latus rectum
latus rectum = 2r2 2(b^2)/a
QED

 

[slugg]

thanks... perfect explanation. Its funny... why didnt i see this? lol

Oh well everyone has off days