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Stuck on a Conic

could you point me in the right direction?
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Reply 1
29 views and nobody helps .... 3 hours wait
appalling service ...
I am clueless on this topic. Sorry I'm a recruit
Original post by TeeEm
29 views and nobody helps .... 3 hours wait
appalling service ...


I am utterly disgraceful at maths but this comment made me laugh a lot. Would have been 10/10 with the appropriate stock image from your bountiful collection :lol:
Reply 4
Original post by iEthan
I am utterly disgraceful at maths but this comment made me laugh a lot. Would have been 10/10 with the appropriate stock image from your bountiful collection :lol:


do not want to over do it ....
Reply 5
Original post by Student403
I am clueless on this topic. Sorry I'm a recruit


ONLY strong FP2/FP3 ....
you know that I have to sign your reference if you go to the States ?
Original post by TeeEm
do not want to over do it ....


After your results at the TSR awards tonight, I think you could afford to let your hair down :tongue: (congratulations by the way, voted for you :yep:)
Original post by TeeEm
ONLY strong FP2/FP3 ....
you know that I have to sign your reference if you go to the States ?


:emo: I will crack down on the two once I have finished them by Easter
Reply 8
Original post by iEthan
After your results at the TSR awards tonight, I think you could afford to let your hair down :tongue: (congratulations by the way, voted for you :yep:)


I have let my hair down ...
I am doing some mechanics questions at the moment as I am very happy!!
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Reply 9
Original post by TeeEm
...


I am nearing a solution, but you will hate me for how inelegant it is.
Original post by TeeEm
I have let my hair down ...
I am doing some mechanics questions at the moment as I am very happy!!
c6.jpg


This image makes me very happy, thank you ^^
Reply 11
Original post by Student403
:emo: I will crack down on the two once I have finished them by Easter


Zacken is also hiding ...
he suddenly remembered he had to shampoo his beard ....
Reply 12
Original post by TeeEm
Zacken is also hiding ...
he suddenly remembered he had to shampoo his beard ....


Original post by Zacken
I am nearing a solution, but you will hate me for how inelegant it is.


Oi! :tongue:
Reply 13
Original post by Zacken
I am nearing a solution, but you will hate me for how inelegant it is.


ready to be impressed
last years under 18 mathmo would have gone to @Gome44


this year still a while to go, you have to earn this year's title against, a strong crowd such as @physicsmaths @Krollo @16Characters.... @Renzhi10122 @B_9710 @Duke Glacia @Louisb19 to name a few
Reply 14
Original post by TeeEm
could you point me in the right direction?
Presentation1.jpg


Please don't kill me, I know it's inelegant! :frown:

Your ellipse is nothing but:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\left(x - \frac{3}{2}\right)^2}{(\frac{5}{2})^2} + \frac{y^2}{2^2} = 1\end{equation*}



We can shift things along using u=x32u = x - \frac{3}{2} so that our foci are at O(3/2,0)O(-3/2, 0) and T(0,3/2)T(0, 3/2).

Now, we know that: PO=ePMPO = ePM' and PT=ePMPT = ePM, but from a simple sketch, we can see:

PM=u+5/2ePM' = u + \frac{5/2}{e} and PM=52euPM = \frac{\frac{5}{2}}{e} - u.

So: OP+PT=e(5/2eu)+e(5/2e+u)=2×52=5|OP| + |PT| = e\left(\frac{5/2}{e} - u\right) + e\left(\frac{5/2}{e} + u\right) = 2\times \frac{5}{2} = 5
(edited 8 years ago)
Reply 15
Original post by Zacken
Please don't kill me, I know it's inelegant! :frown:

Your ellipse is nothing but:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\left(x - \frac{3}{2}\right)^2}{(\frac{5}{2})^2} + \frac{y^2}{2^2} = 1\end{equation*}



We can shift things along using u=x32u = x - \frac{3}{2} so that our foci are at O(3/2,0)O(-3/2, 0) and T(0,3/2)T(0, 3/2).

Now, we know that: PO=ePMPO = ePM' and PT=ePMPT = ePM, but from a simple sketch, we can see:

PM=u+5/2ePM' = u + \frac{5/2}{e} and PM=52euPM = \frac{\frac{5}{2}}{e} - u.

So: OP+PT=e(5/2eu)+e(5/2e+u)=2×52=2|OP| + |PT| = e\left(\frac{5/2}{e} - u\right) + e\left(\frac{5/2}{e} + u\right) = 2\times \frac{5}{2} = 2


I think the method is fine but I think there is a number error somewhere
(you do not post you workings instead you show off you LaTex, so I cannot locate it)
Reply 16
Original post by TeeEm
I think the method is fine but I think there is a number error somewhere
(you do not post you workings instead you show off you LaTex, so I cannot locate it)


Is the form my ellipse correct?
Reply 17
Original post by Zacken
Is the form my ellipse correct?


I do not think so
Reply 18
Original post by TeeEm
I do not think so


I did: 5r3x=825r2=(8+3x)225(x2+y2)=64+48x+9x2\displaystyle 5r - 3x = 8 \Rightarrow 25r^2 = (8+3x)^2 \Rightarrow 25(x^2 + y^2) = 64 + 48x + 9x^2, so re-arranging:

16x248x+25y2=6416((x3/2)29/4)+25y2=64\displaystyle 16x^2 - 48x + 25y^2 = 64 \Rightarrow 16((x-3/2)^2 - 9/4) + 25y^2 = 64

So 16(x3/2)236+25y2=64\displaystyle 16(x-3/2)^2 - 36 + 25y^2 = 64 Hence:

16(x3/2)2+25y2=10016(x3/2)2100+y24=1\displaystyle 16(x-3/2)^2 + 25y^2 = 100 \Rightarrow \frac{16(x-3/2)^2}{100} + \frac{y^2}{4} = 1
Reply 19
Original post by Zacken
I did: 5r3x=825r2=(8+3x)225(x2+y2)=64+48x+9x2\displaystyle 5r - 3x = 8 \Rightarrow 25r^2 = (8+3x)^2 \Rightarrow 25(x^2 + y^2) = 64 + 48x + 9x^2, so re-arranging:

16x248x+25y2=6416((x3/2)29/4)+25y2=64\displaystyle 16x^2 - 48x + 25y^2 = 64 \Rightarrow 16((x-3/2)^2 - 9/4) + 25y^2 = 64

So 16(x3/2)236+25y2=64\displaystyle 16(x-3/2)^2 - 36 + 25y^2 = 64 Hence:

16(x3/2)2+25y2=10016(x3/2)2100+y24=1\displaystyle 16(x-3/2)^2 + 25y^2 = 100 \Rightarrow \frac{16(x-3/2)^2}{100} + \frac{y^2}{4} = 1


correct now
:smile:

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