Easy qs from JR Trial (2 Viewers)

lychnobity · Oct 10, 2009

Brain malfunctioning atm. Someone please help.

ninetypercent · Oct 10, 2009

bi) For tangent at P, y = tx - at^2
For tangent at Q, y = ux - au^2
solving simultaneously
tx - at^2 = ux - au^2
tx - ux = -au^2 + at^2
x(t-u) = -a(q^2 - t^2)
x(t-u) = -a (u-t) (u+t)
x = a (u+t)

substitute x in y = ux - au^2
y = ua(u+t) - aq^2
= u^2a + uat - au^2
= uat

the tangents intersect at (a(u+t), uat)

lychnobity · Oct 10, 2009

ninetypercent said:
bi) For tangent at P, y = tx - at^2
For tangent at Q, y = ux - au^2
solving simultaneously
tx - at^2 = ux - au^2
tx - ux = -au^2 + at^2
x(t-u) = -a(q^2 - t^2)
x(t-u) = -a (u-t) (u+t)
x = a (u+t)

substitute x in y = ux - au^2
y = ua(u+t) - aq^2
= u^2a + uat - au^2
= uat

the tangents intersect at (a(u+t), uat)

Thanks, but I got that part. I'm more interested in the second part (which I'm having trouble with)

Trebla · Oct 10, 2009

For the second one:

$(5+2x)^{12}=\sum_{k=0}^{12}\binom{12}{k}5^{12-k}(2x)^{k}\\\\= \sum_{k=0}^{12}\binom{12}{k}5^{12-k}2^kx^k\\\\\therefore a_k=\binom{12}{k}5^{12-k}2^k\\\\\dfrac{a_{k+1}}{a_k}=\dfrac{\binom{12}{k+1}5^{11-k}2^{k+1}}{\binom{12}{k}5^{12-k}2^k}\\\\= \dfrac{2\binom{12}{k+1}}{5\binom{12}{k}}\\\\=\dfrac{2}{5}\dfrac{12!}{(k+1)!(11-k)!}\dfrac{k!(12-k)!}{12!}\\\\=\dfrac{2}{5}\dfrac{k!(12-k)(11-k)!}{(k+1)k!(11-k)!}\\\\=\dfrac{2}{5}\dfrac{(12-k)}{(k+1)}\\\\=\dfrac{24-2k}{5k+5}$

lychnobity · Oct 10, 2009

thanks trebla

hermand · Oct 10, 2009

b. i.
$tx-at^{2}=ux-au^{2}$
$tx-ux=at^{2}-au^{2}$
$x(t-u)=a(t-u)(t+u)$
$x=\frac{a(t-u)(t+u)}{(t-u)}$
$x=a(t+u)$
$\therefore y=t[a(t+u)]-at^{2}$
$y=at^{2}+atu-at^{2}$
$y=atu$
$\therefore \textup{intersection point}(a(t+u),atu)$

hermand · Oct 10, 2009

aww it took me so long to do that on eqn editor. if you still don't have an answer for it in the morning i'll help out, but it is way too far past my bedtime.

hermand · Oct 10, 2009

umm second part, t&u are equal to -2 and 3, then show by distance formula that two sides are equal. my head is hurting too much to type the whole solution out.

lychnobity · Oct 10, 2009

hermand said:
aww it took me so long to do that on eqn editor. if you still don't have an answer for it in the morning i'll help out, but it is way too far past my bedtime.

lol but thanks for taking the time to help out, i appreciate it.

lychnobity · Oct 10, 2009

hermand said:
umm second part, t&u are equal to -2 and 3, then show by distance formula that two sides are equal. my head is hurting too much to type the whole solution out.

umm do you think your head will be ok if i asked you to just run through the method?

edit: nevermind, saw it now.

hermand · Oct 10, 2009

sorry, i fell straight to sleep almost with my computer on my lap. i still had notes and books everywhere on my bed. haha.

sorry!

Top Secret · Oct 10, 2009

hermand said:
sorry, i fell straight to sleep almost with my computer on my lap. i still had notes and books everywhere on my bed. haha.

sorry!

Haha, I do that all the time!

lychnobity · Oct 10, 2009

hermand said:
sorry, i fell straight to sleep almost with my computer on my lap. i still had notes and books everywhere on my bed. haha.

sorry!

haha, not to worry, I 'saw' it as soon as I posted it anyhow.

lychnobity · Oct 10, 2009

Stuck again, help

clintmyster · Oct 10, 2009

lychnobity said:
Stuck again, help

i) angle COQ = angle TQB = 90 (data)
therefore APTQ is a cyclic quad (exterior angle of cyclic quad = interior opposite angle)
therefore angle TAQ = angle TPQ (angles standing on the same arc = )
CPQB is a cyclic quad (CPB and CQB are semicircles with diameter BC)
therefore angle BPQ = angle TPQ = angle BCQ (angles standing on same arc =)
therefore angle TAQ = angle QCB

ii) Therefore AQRC is a cyclic quadrilateral (angles on the same are = )
Therefore angle TQC = angle ARC = 90 (angles on the same arc = )
Thus AR perpendicular to CB

Knowing latex would be nice but cbf learning haha. Hope that helps

lychnobity · Oct 10, 2009

Thanks man.

um, to use latex, click the quick reply button (see the 'quote' button? count 2 to the right), then go to the bottom of the page and click 'launch latex equation editor' and click around to write up the solution. When you're done click 'copy to document' and submit the post.

lychnobity · Oct 10, 2009

clintmyster said:
i) angle COQ = angle TQB = 90 (data)
therefore APTQ is a cyclic quad (exterior angle of cyclic quad = interior opposite angle)
therefore angle TAQ = angle TPQ (angles standing on the same arc = )
CPQB is a cyclic quad (CPB and CQB are semicircles with diameter BC)
therefore angle BPQ = angle TPQ = angle BCQ (angles standing on same arc =)
therefore angle TAQ = angle QCB

ii) Therefore AQRC is a cyclic quadrilateral (angles on the same are = )
Therefore angle TQC = angle ARC = 90 (angles on the same arc = )
Thus AR perpendicular to CB

Knowing latex would be nice but cbf learning haha. Hope that helps

Sorry to bother but where's angle COQ?

Top Secret · Oct 10, 2009

lychnobity said:
Sorry to bother but where's angle COQ?

I think he means angle APT.

Drongoski · Oct 10, 2009

Top Secret said:
I think he means angle APT.

Whatever that is: APTQ is cyclic because opposite angles supplementary:

i.e. angl APT (= 90 deg) + angl AQT (90 deg) = 180 deg

Drongoski · Oct 10, 2009

clintmyster said:
i) angle COQ = angle TQB = 90 (data)
therefore APTQ is a cyclic quad (exterior angle of cyclic quad = interior opposite angle)
therefore angle TAQ = angle TPQ (angles standing on the same arc = )
CPQB is a cyclic quad (CPB and CQB are semicircles with diameter BC)
therefore angle BPQ = angle TPQ = angle BCQ (angles standing on same arc =)
therefore angle TAQ = angle QCB

ii) Therefore AQRC is a cyclic quadrilateral (angles on the same are = )
Therefore angle TQC = angle ARC = 90 (angles on the same arc = )
Thus AR perpendicular to CB

Knowing latex would be nice but cbf learning haha. Hope that helps

Good work !

Easy qs from JR Trial (2 Viewers)

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