Weird conics answer (1 Viewer)

[sasquatch]

Ok for this question it states:

Sketch showing all major features:

(x-2)²/4 - (y+1)²/16 = 1


so please look closely and tell me if i have done it right

i say b = 4, a = 2,

and hence 16 = 4(e² - 1)
thefore e = root 5

yep so if thats alright then my answers right.

but for some reason the asnwer in the back of terry lee has it the other way around. I know that for ellipses, the major axis length has to be greater than the minor axis length. But this does not apply to rectangular hyperbolae.

For other questions with b > a, he does not flip them around, so is this a simple mistake.. cuz now im getting really confused...

 

[sasquatch]

Wait nevermind, i saw that in my graph the curve cuts the y-axis at (0,-1) but terry lee's is different.. So i checked it on a graphing program...and i was right!! terry lee is wrong


WRONG!!!!!

 

[Riviet]

Just curious, but which edition of Terry Lee's is this?

 

[sasquatch]

hehe you've asked me that question twice before in two other posts. Its 5th edition the black one. Its old but it seemed to have more questions or something, cuz it had like multiple exercises like A, B, C for the same set of work. Also i swear to god the printing in the new one sucks so much... its too blue and the blues crap and blotchy. AND the black text on those dark blue blackground sucks to read.. cheap printing...

 

[Riviet]

Yeah sorry, I forgot, also I gotta agree with your point on the printing in the 6th edition. On some pages, the question number would be left out so it would be really confusing when doing those set of questions. I don't mind the blueness though, it's my favourite colour. XD

 

[sasquatch]

Yeah i like blue too.. but that blues just yuck.

 

[pLuvia]

In hyperbolas the it doesn't matter if the b or a is bigger, but yes I've encountered those questions from him as well, I just tend to use b²>a²

 

[sasquatch]

Instead of making another thread... id like to ask another question.

I didnt really understand terry lees explinations of the deriveration of the parametric equations for both the ellipse and rectangular hyperbola... i swear its like he wants people to come to his tutoring to find out or something... theres hardly an explination in the book.. but anyway. My teacher gave us this coroneos book which has conics stuff in it which is so much better for explinations. I completely understand how the parametric equations come from the rectangular hyperbola, but still confused with the ellipse.

Consider the graph i attached.

View attachment 12852

In Coroneos' book, he says PN/QN = b. How is that so.. from that picture (same as he provides) i cannot see how that is true. And also using this factor, i still dont get how to get the set for y. The x-value is the same i see so, x = acos@

PN/QN = b,

but QN = asin@

so PN = bQN,
PN = absin@

and isnt PN the y value for the point, so thats confusing!!! It would fit if it PN/QN = b/a, but i dunno...

Could someone please help me out?

 

[pLuvia]

The circle is an auxiliary circle x²+y²=a² so the parametric equation is (acosθ,asinθ). So PN=bsinθ, QN=asinθ, PN/QN=b/a. To get the y-value for the ellipse, you just sub x=acosθ into the ellipse general equation and you should yield bsinθ

So I'm guessing it's just a typo

 

[sasquatch]

Yeah i was thinking it was a typo... but i just wasnt sure cuz it doesnt go any further than that, well it says sin@/asin@ = b/a...but i dont get what that means...

Um yeah why is PN/QN = b/a though?

 

[Riviet]

It's what pLuvia just said in the last post. Notice that P and Q have the same x-value, acosθ, so co-ordinates of P are (acosθ,bsinθ) and co-ordinates of Q are (acosθ,asinθ) PN refers to the y co-ordinate of P and QN refers to the y co-ordinate of Q.

.'. PN/QN = bsinθ/asinθ = b/a

 

[sasquatch]

Yes yes... i get all of that. Your statement though,

.'. PN/QN = bsinθ/asinθ = b/a

is working backwards.. I get the x-cordinates are the same for both the ellipse and its auxiliary circle.

This question is to derive the parametric equation of the ellipse. so we must find PN (the y-value for point Q) oh shit i just realised in my diagram i put the P and Q the wrong way... yeah but sorry.. So my question is why is PN/QN = b/a without working backwards like you did as in stating,

QN = y = asin@

PN = b/a * QN
= b/a * asin@
= bsin@

Get?

 

[Riviet]

Let O be origin. Then let OQ make an angle θ with the positive direction of the x-axis, we'll use your diagram provided of course, so θ is acute. We know the OP has length of a, so using simple right triangle trig ratios, Q has co-ordinates (acosθ,asinθ). We know that general equation of ellipse is x²/a² + y²/b² = 1 and P and Q have a common x co-ordinate of acosθ so substitute it into equation of ellipse to obtain y=bsinθ.

.'. ellipse has parametric equations (acosθ,bsinθ).

I hope that's what you were asking for.

 

[sasquatch]

Yeah!!! Thats what i wanted.. hmm i was thinking cuz coroneous had that PN/QN =b/a that you had to use that to derive the y-part of the parametric equation. But now i see! Thanks alot man...

 

[Riviet]

Woot! Nice work, pleasure to help.