Author Topic: Mathematics Question Thread (Read 1626121 times) Tweet Share

Justinhales · « **Reply #3090 on:** October 23, 2017, 08:41:58 am »

Hey!!
Sorry real quick question..!!

A parabola has focus (5, 0) and directrix x = 1.
What is the equation of the parabola?

Can u pls do full working out too!!
thanks!

RuiAce · « **Reply #3091 on:** October 23, 2017, 08:47:16 am »

Quote from: Justinhales on October 23, 2017, 08:41:58 am

Hey!!
Sorry real quick question..!!

A parabola has focus (5, 0) and directrix x = 1.
What is the equation of the parabola?

Can u pls do full working out too!!
thanks!

$\text{Use a sketch.}$
Nobody really cares if the sketch is rough or not, but you need to sketch it.

$\text{Since the vertex is halfway between the directrix and focus}\\ \text{it is at }(3,0)$
$\text{Also the focal length is the distance from the focus to the vertex}\\ \text{which is }a=2$
$\text{So as we have a 'concave-right' parabola}\\ \text{its equation is }(y-k)^2=4a(x-h)\\ \text{so }\boxed{y^2=8(x-3)}$
There's not much working out at all for these questions.

Justinhales · « **Reply #3092 on:** October 23, 2017, 08:53:25 am »

Ta!!!

angelahchan · « **Reply #3093 on:** October 23, 2017, 09:30:44 am »

Hi, can someone please help with Q9 of 2016? I'm not sure how to integrate with absolute values
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF

winstondarmawan · « **Reply #3094 on:** October 23, 2017, 09:32:53 am »

Are we permitted to use this formula:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809677_1357774924348027_1303197799_n.jpg?oh=76d0e9125ddfa8d3f76ff4dbe96fe3ed&oe=59EF9CD2
Even though it's not in the syllabus?

Opengangs · « **Reply #3095 on:** October 23, 2017, 09:36:15 am »

Quote from: angelahchan on October 23, 2017, 09:30:44 am

Hi, can someone please help with Q9 of 2016? I'm not sure how to integrate with absolute values
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF

Recall that the integral is merely the area under the curve.
Sketch the absolute function to find the area of the two triangles formed between the two bounds.
If you're still unsure, let me know!!

Quote from: winstondarmawan on October 23, 2017, 09:32:53 am

Are we permitted to use this formula:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809677_1357774924348027_1303197799_n.jpg?oh=76d0e9125ddfa8d3f76ff4dbe96fe3ed&oe=59EF9CD2
Even though it's not in the syllabus?

I'm pretty sure then you have to derive it, if it's not in the syllabus

pokemonlv10 · « **Reply #3096 on:** October 23, 2017, 09:39:22 am »

how would you do this question?

mbdtHSC · « **Reply #3097 on:** October 23, 2017, 10:13:56 am »

Quote from: pokemonlv10 on October 23, 2017, 09:39:22 am

how would you do this question?

I think you split it into an AP with 7, 14, 21... and GP with 3, 6, 12 ... then sum each

Opengangs · « **Reply #3098 on:** October 23, 2017, 10:15:49 am »

Quote from: pokemonlv10 on October 23, 2017, 09:39:22 am

how would you do this question?

$\text{Recall that the formula for the sum to n terms of an AP is: } A_n = \frac{n}{2}[2a + (n - 1)d]$
$\text{Recall that the formula for the sum to n terms of a GP is: } G_n = \frac{a(r^n - 1)}{r - 1}$

If we split up the series, we can form something like:
$\underbrace{7 + 14 + 21 + ...}_{AP} + \underbrace{3 + 6 + 12 + ...}_{GP}$
$\text{Every odd term we have belongs to the AP, while every even term belongs to the GP}$
$\therefore \underbrace{A_1 + A_3 + A_5 + ... + A_{51}}_{AP} + \underbrace{A_2 + A_4 + A_6 + .. + A_{50}}_{GP}$
$\therefore A_{26} = 13(14 + 175) = 2457$
$G_{25} = \frac{3(2^{25} - 1)}{1} = 3(2^{25} - 1)$

$\therefore A_{26} + G_{25} = 2457 + 3(2^{25} - 1) = 100665750$

pokemonlv10 · « **Reply #3099 on:** October 23, 2017, 10:39:16 am »

Not sure about this trig

RuiAce · « **Reply #3100 on:** October 23, 2017, 10:43:19 am »

Quote from: pokemonlv10 on October 23, 2017, 10:39:16 am

Not sure about this trig

$\text{Recall that }\sin (-x) = -\sin x\text{ and }\sin \left(\frac\pi 2 - x\right) = \cos x\\ \text{This gives }\boxed{\sin \left(\theta - \frac\pi2\right) = -\cos \theta}$
$\text{Also recall that }\cos \left(\frac\pi2 - x\right)= \sin x\\ \text{Hence }\boxed{\cos \left(\frac\pi2 - (-\theta)\right) = \sin (-\theta) = -\sin \theta}$
$\text{So ultimately, we are solving the equation}\\ \begin{align*}-\cos \theta &= -\sin \theta\\ \implies 1&=\tan \theta\end{align*}\\ \text{which will have solutions }\theta=\frac\pi4, \frac{5\pi}4$

winstondarmawan · « **Reply #3101 on:** October 23, 2017, 10:59:56 am »

Hello!
Just a little confused on how they got 5pi4 out.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22790869_816333725195103_370366407_o.png?oh=65e3b77173cc2771d0f8b9c4ab7bbf0a&oe=59EF5BC6
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809895_816336158528193_1950803996_n.jpg?oh=c5927f4ef4917a4b2cbce9df04d4e20f&oe=59EF96A1
TIA.

RuiAce · « **Reply #3102 on:** October 23, 2017, 11:02:57 am »

Quote from: winstondarmawan on October 23, 2017, 10:59:56 am

Hello!
Just a little confused on how they got 5pi4 out.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22790869_816333725195103_370366407_o.png?oh=65e3b77173cc2771d0f8b9c4ab7bbf0a&oe=59EF5BC6
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809895_816336158528193_1950803996_n.jpg?oh=c5927f4ef4917a4b2cbce9df04d4e20f&oe=59EF96A1
TIA.

They didn't really adjust the domains like they should have. Which is fair enough, because adjusting the domains would be dodgy (since we need to overlap the domains, and THEN work out the relevant quadrants). But I don't really like that method because it can be too confusing to explain.

Just use my method above.

caitlinlddouglas · « **Reply #3103 on:** October 23, 2017, 11:04:34 am »

They could someone please explain the solution to 15c) ii please? Especially why there are the less than and greater than/equal to signs Thanks!

RuiAce · « **Reply #3104 on:** October 23, 2017, 11:05:25 am »

Quote from: caitlinlddouglas on October 23, 2017, 11:04:34 am

They could someone please explain the solution to 15c) ii please? Especially why there are the less than and greater than/equal to signs Thanks!

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