Print Page - Holiday Homework Help!

Title: Holiday Homework Help!
Post by: Batwing on January 28, 2013, 03:35:03 pm

Hey everyone. I really need help with this loci question! I shouldn't have left it until now!

Title: Re: Holiday Homework Help!
Post by: satya on January 29, 2013, 12:25:01 pm

Quote from: Batwing on January 28, 2013, 03:35:03 pm

Hey everyone. I really need help with this loci question! I shouldn't have left it until now!

by any chance would this be a CAS question or?

Title: Re: Holiday Homework Help!
Post by: polar on January 29, 2013, 03:33:23 pm

I've put all the working in spoiler tags because it's quite long, I guess you could use the stuff not in spoiler tags as hints if you want
first you need to find the general formula for the normal at any value of $0 \le \theta \le \frac{\pi}{2}$ . so find the gradient of the normal in terms of $\theta$ ,

Spoiler

$\begin{aligned} x^2 + \frac{y^2}{b^2} &= 1 \\ y&=b\sqrt{1-x^2} \; \; (\text{Take positive since it's in quadrant 1}) \\ \frac{dy}{dx} &= b\times\frac{-x}{\sqrt{1-x^2}} \\ \text{Substitute \;} x=\cos(\theta), y=\sin(\theta) \\ \frac{dy}{dx} &= b\times\frac{-\cos(\theta)}{\sqrt{1-\cos^2(\theta)}} \\ &= b\times\frac{\cos(\theta)}{\sin(\theta)} \\ -\frac{dx}{dy} &= \frac{\tan(\theta)}{b}\end{aligned}$

next, substitute into $y-y_1 =m(x-x_1)$ ,

Spoiler

$y-b\sin(\theta) = \frac{\tan(\theta)}{b}(x-\cos(\theta))$
find the y-intercept by letting x=0
$\begin{aligned} y - b\sin(\theta) &= \frac{\tan(\theta)}{b}(0-\cos(\theta)) \\ y &= b\sin(\theta) - \frac{\sin(\theta)}{b} \\ y &= \frac{b^2-1}{b}\sin(\theta) \end{aligned}$

hence, N is $\left(0,\frac{b^2-1}{b}\sin(\theta)\right)$

now, you can find M, the midpoint of N and P

Spoiler

$M\left(\frac{\cos(\theta)}{2}, \frac{2b^2-1}{2b}\sin(\theta)\right)$
now, we know that
$x=\frac{\cos(\theta)}{2}, y=\frac{2b^2-1}{2b}\sin(\theta)$
solve for $\cos(\theta)$ and $\sin(\theta)$
$\cos(\theta) = 2x, \sin(\theta) = \frac{2b}{2b^2-1}y$

hence, the locus is $4x^2 + \left(\frac{2b}{2b^2-1}\right)^2y^2 = 1$ , but if $x=\frac{1}{\sqrt{2}}$ , it's undefined.

the locus is a circle if the coefficient of $x^2$ is the same as $y^2$

Spoiler

hence, solve $\left(\frac{2b}{2b^2-1}\right)^2=4$ which gives $b=\pm1, \pm\frac{1}{2}$

Title: Re: Holiday Homework Help!
Post by: Batwing on January 29, 2013, 09:51:55 pm

Thank you so much for the help! I just couldn't figure out the last part. You are a lifesaver.

Title: Re: Holiday Homework Help!
Post by: barydos on February 04, 2013, 05:29:47 pm

Oh my, I've got the same work set this week, and this was the question that I couldn't do! Thanks so much for asking and for the answer :D

EDIT: Sorry, how do you do part e) ii)? I'm quite confused there.
Subbing in $b =$ $\frac{1}{\sqrt{2}}$ gives $4x^2 + undefined = 1$
or.. maybe I did something wrong?

ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Batwing on January 28, 2013, 03:35:03 pm