Lactus Rectum (1 Viewer)

BillyMak

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The latus rectum is the line that goes through the focus and is perpendicular to the axes that the focus is on. It is bounded by the two points where it intersects the curve, and for a parabola it's length is 4a. I can't remember off the top of my head what the corresponding lengths are for the hyperbola and ellipse, but it isn't that hard to figure out.
 

mojako

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tywebb said:
latus latus latus latus latus?

yes.
lactus?

no.

ok?
hmm...
latus sounds like a flower to me.....
laCtus sound like caCtus....

:headbang: :bat: :gridnod: :cool: :argue: :santa: :angry:

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The images don't even come up :angry:
 
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Slidey

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I think it is the two lines perpendicular to the major axis, both lines passing through a focus, in an ellipse. In a circle it's just the diameter.

Oh and "lactus rectum" sounds like "milky posterior".
 
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BillyMak

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Could you say that a ellipse perhaps has 3 latus recta?

I mean consider the basic ellipse, (x<sup>2</sup>/a<sup>2</sup>) + (y<sup>2</sup>/b<sup>2</sup>) = 1, where a > b, you could say that there are the two vertical latus recta as well as one horizontal latus rectum that goes through both foci?
 

BillyMak

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Meh. My mistake, forgot the requirement that the focus has to be the midpoint of the line, or it isn't a latus rectum.
 

mojako

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oh, sir,
I thought u meant this as a joke.. a really funny one
latus latus latus latus latus?
laCtus?

didn't realise that mazza spelt it wrong

tywebb said:
latus latus latus latus latus?

yes.
laCtus?

no.

ok?
 

Slidey

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BillyMak said:
Could you say that a ellipse perhaps has 3 latus recta?

I mean consider the basic ellipse, (x<sup>2</sup>/a<sup>2</sup>) + (y<sup>2</sup>/b<sup>2</sup>) = 1, where a > b, you could say that there are the two vertical latus recta as well as one horizontal latus rectum that goes through both foci?
Well, I was thinking of those two lines, but tywebb doesn't seem to think so, so...
 

jumb

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so for the ellipse (x2/a2) + (y2/b2) = 1, where a > b, the latus rectum is x = +/- ae ?
 

nick1048

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mojako said:
hmm...
latus sounds like a flower to me.....
laCtus sound like caCtus....

:headbang: :bat: :gridnod: :cool: :argue: :santa: :angry:

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The images don't even come up :angry:[/QUOTE]

lol a flower hahahaha lotus is a flower and a very old car racing game. lactus sounds like lactic acid, ouch exercise pain!
 

BillyMak

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jumb said:
so for the ellipse (x2/a2) + (y2/b2) = 1, where a > b, the latus rectum is x = +/- ae ?
Yes. (ladidadida there's 10 characters)
 

withoutaface

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BillyMak said:
Could you say that a ellipse perhaps has 3 latus recta?

I mean consider the basic ellipse, (x<sup>2</sup>/a<sup>2</sup>) + (y<sup>2</sup>/b<sup>2</sup>) = 1, where a > b, you could say that there are the two vertical latus recta as well as one horizontal latus rectum that goes through both foci?
A latus rectum has to be parallel to a directrix.
 

Premus

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So its just two lines passing through each foci and parallel to the directrix?
and what would be the length?

thanks
 

BillyMak

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PremusDog said:
and what would be the length?

thanks
For a parabola, x<sup>2</sup> = 4ay, the y value is a at the points where the latus rectum cuts the curve.

So x<sup>2</sup> = 4a<sup>2</sup>
x = +/- 2a
Therefore length of latus rectum for parabola is 2*2a = 4a

For an ellipse, (x<sup>2</sup>/a<sup>2</sup>) + (y<sup>2</sup>/b<sup>2</sup>) = 1, where a>b, the x-value of the latus recta where it cuts the curve are +/- ae,
When x = ae:

(a<sup>2</sup>e<sup>2</sup>/a<sup>2</sup>) + (y<sup>2</sup>/b<sup>2</sup>) = 1

y<sup>2</sup> = b<sup>2</sup>(1 - e<sup>2</sup>)
y = +/- b(1- e<sup>2</sup>)<sup>1/2</sup>

so length of latus recta is 2b(1- e<sup>2</sup>)<sup>1/2</sup>
 

BillyMak

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For a hyperbola, couldn't be bothered doing the working but it would be 2b(e<sup>2</sup> - 1)<sup>1/2</sup>
 

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