We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+10 **

The asymptotes will go through the center of the hyperbola, which is (-2,5).

The square roots of the denominators of the x-term and the y-term give the slope of the lines. Since slope is y-value over x-value, the slopes will be 9/11 and -9/11.

The equations will be y - 5 = ± (9/11)(x + 2).

When graphing, you draw the box; the asymptotes are the lines created by the opposite corners of the box.

geno3141 Aug 15, 2015

#3**+5 **

These graphs are always a nice addition to the answers.........thanks, Melody

CPhill Aug 15, 2015

#4**0 **

I am just playing a little with this question and looking at Gino's answer.

I am not familiar with this topic.

This quiet froum is great I have time to go back and look at some of the old questions and answers

I need to find the asymptotes of this hyperbola (x+2)^2/121-(y-5)^2/81=1

\(\frac{(x+2)^2}{121}-\frac{(y-5)^2}{81}=1\\ centre\;\;(-2,5)\\\)

The square roots of the denominators of the x-term and the y-term give the slope of the lines. Since slope is y-value over x-value, the slopes will be 9/11 and -9/1

\(gradients = \frac{\pm \sqrt{y\; denominator}}{\sqrt{x\; denominator}}=\frac{\pm \sqrt{81}}{\sqrt{121}}=\pm \frac{ 9}{11}\)

Why?

The equations will be y - 5 = ± (9/11)(x + 2).

When graphing, you draw the box; the asymptotes are the lines created by the opposite corners of the box. ???

Mmm Interesting....... More thought needed .......

Melody Dec 29, 2015

#6**0 **

Thanks Alan, yes I do find that very interesting.

I could follow all your working through, it was very clear.

I still want to spend more time processing though.

There are concepts that I need to cement into my head here.

Melody Dec 29, 2015

#7**+5 **

Hi Melody.

You probably have this sorted by now, but if you haven't here's the usual approach.

Begin with a change of variables, let

\(\displaystyle X=x+2\qquad Y=y-5\)

so

\(\displaystyle \frac{X^{2}}{121}-\frac{Y^{2}}{81}=1\) ,

which in the limit (sort of) as \(X\) and \(Y\) go to infinity becomes

\(\displaystyle Y = \pm \frac{9}{11}X\) .

Those are the asymptotes relative to the \(\displaystyle X\) and \(Y\) axes.

Now switch back to the original co-ordinate system,

\(\displaystyle y-5 = \pm \frac{9}{11}(x+2)\) .

Guest Dec 30, 2015

#8**0 **

Thanks guest, I would be very interested in what you have to say but your post will not display properly.

It did display for about 2 seconds but then it disappeard again.

Melody Dec 30, 2015

#9**0 **

Thanks Guest, it is displaying now and I looked at what you presented.

Yes I like that, I'd like some more to them to play with though. ://

Melody Dec 30, 2015

#10**+5 **

I'm not quite sure what you're looking for, but try this one.

For the given hyperbola, (though the result is true for any hyperbola, prove the general result if you want), show that if the tangent at any point P cuts the asymptotes at Q1 and Q2, then P is the mid-point of the line joining Q1 to Q2.

Guest Dec 30, 2015

#11**0 **

Thanks, I have been looking at your question but I have not exhausted it yet.

Would you please consider becoming a member, or if you do not want to, would you consider attaching a name to your posts. I am please that you are helping me but I'd really like you to have an identity that I can recognise and be very pleased to see. :)

Melody Dec 31, 2015

#12**0 **

It’s me. Your friendly, neighborhood’s troll :)

I thought to get a head start on my new year’s resolution to be more helpful.

I have to tell you though, It's not even the new year yet, and I know I will fail:

It’s making me more nauseated.

Happy New Year :)

Guest Dec 31, 2015

#13**0 **

Hi Naus,

I know your message is intended for our guest but that won't stop me from adding my 2 cents worth!

I have a tip for you.

When you are posting incognito, it is best not to give your name......

It kinda gives your identity away!

Anyway, **Happy New Year** to you.

I even found you a gift!

[You just need to extract it from the previous owner first - No bid deal for you I am sure ]

Melody Dec 31, 2015

#14**+5 **

Bertie

Sorry Melody, it was me, I'm responsible for both #7 and #10, I just forgot to sign them.

The login problem has not been resolved. I login and my name appears up in the top right hand corner and also one of those thingys showing who's on line appears at the bottom. Then as soon as I try to do something, like reading mail, I'm told that I'm not logged in. It would appear that I am not alone, a number of others seem to have the same problem.

Anyway, I hope that Christmas went well for you, and really best wishes for the new year

Bertie

Guest Dec 31, 2015

#15**+5 **

I knew it was you Bertie; I recognized your voice and your accent. Happy New Year.

Subtly signing my name to your work was a great way to troll Melody for the start of the New Year.

Thanks Melody for the Paddywhack and giving this troll a bone. Rover doesn’t mind sharing.

Nauseated Dec 31, 2015

#16**0 **

Hahaha, I had to read your message about a dozen times before i realized what you were talking about Naus.

I never even considered that anon might be you!! I knew that anon was both smart and sweet so he could not have been you!! ROF LOL

Thanks Bertie, it is always a great pleasure having you on the forum. You must remember to sign your name at the beginning of our posts though!

**A Very Happy New Year to you both. **

Melody Jan 1, 2016