Author Topic: 4U Maths Question Thread (Read 664944 times) Tweet Share

Rathin · « **Reply #795 on:** February 03, 2017, 04:28:31 pm »

Write the equation of the locus at point P that moves such that its distance from (3,0) is 4/5 as its distance from the line x=2.

RuiAce · « **Reply #796 on:** February 03, 2017, 04:45:41 pm »

Quote from: Rathin on February 03, 2017, 04:28:31 pm

Write the equation of the locus at point P that moves such that its distance from (3,0) is 4/5 as its distance from the line x=2.

$\textit{This question aims to find the equation of an ellipse from first principles.}$
$\text{Denote }P\text{ as the variable point }(x,y).\\ \text{Let }S\text{ be the point }(3,0)\\ \text{and let }M\text{ be the point }(2,y)\\ \text{so that }PM\text{ represents the perpendicular distance from }P\text{ to }M.$
$\text{Then, }PS=\frac{4}{5}PM\\ \text{So working off our methods in 2U prelim}$
$\begin{align*}25PS^2&=16PM^2\\ 25[(x-3)^2+y^2]&=16[(x-2)^2+0^2]\\ 25x^2-150x+225+25y^2&=16x^2-64x+64\\9x^2-86x+25y^2&=-161 \\ 9\left(x-\frac{43}{9}\right)^2+25y^2&=161+\frac{1849}{9}\\ 9\left(x-\frac{43}{9}\right)^2+25y^2 &=\frac{400}{9}\end{align*}$
You can rearrange however you want.

armtistic · « **Reply #797 on:** February 03, 2017, 08:52:42 pm »

Quote from: RuiAce on February 01, 2017, 05:54:22 pm

$\textbf{You cannot possibly do this question without a diagram.}$
(Image removed from quote.)
$\text{It is easy to prove that the tangent at }P\text{ is given}\\ x+p^2y=2cp\\ \text{and the gradient of this line is }-\frac{1}{p^2}$
$\text{Hence it can be easily checked that}\\ A\text{ is the point }(2cp,0)\\ B\text{ is the point }\left(0, \frac{2c}{p}\right)$
$\text{Now, supposedly }Q\text{ is the point }\left(cq, \frac{c}{q}\right)\\ \text{But we know that the tangent at }Q\text{ is }\textbf{parallel}\text{ to the tangent at }p\\ \text{It is easy to check that the tangent at }Q\text{ has gradient }-\frac{1}{q^2}$
$\begin{align*}\therefore -\frac{1}{p^2}&=-\frac{1}{q^2}\\ \therefore q&=-p\end{align*}\\ \text{Note that }q\neq p\text{ or }P\text{ and }Q\text{ coincide.}$
$\text{So in fact }Q\text{ is the point }\left(-2cp, -\frac{2c}{p}\right)\\ \text{and the tangent at }Q\text{ has equation }x+p^2y=-2cp$
$\text{Hence }C\text{ is }\left(2cp,0\right)\\ D\text{ is }\left(0,-\frac{2c}{p}\right)$
______________________________
$\textbf{The diagram is extremely useful as clearly, by symmetry}\\ ABCD\text{ is a }\textbf{rhombus}\\ \text{It should be easy to calculate the area now by formula.}$
______________________________
$\text{Note that, the diagram also shows us that }\\ Q\text{ and }P\text{ are points obtained by rotational symmetry about the origin}\\ \text{This is because the coordinates of }Q\text{ are the exact NEGATIVES}\, \text{of that of }P\\ \text{So verily, }O\text{ is just the MIDPOINT of }QP$
$\text{This leads to something important:}\\ QOP\text{ is a straight line AND is the DIAMETER of the circle.}\\ \text{Hence }\angle QRP=\angle QSP=\frac{\pi}{2}\text{ as they are the angles in a semi-circle}$
$\text{So using }m_1m_2=-1\text{ for perpendicularity}\\ \begin{align*}m_{PR}m_{QR}&=-1\\ \frac{\frac{c}{p}-\frac{c}{r}}{cp-cr}\cdot \frac{-\frac{c}{p}-\frac{c}{r}}{-cp-cr}&=-1\\ \frac{\frac{r-p}{pr}}{p-r}\cdot \frac{\frac{r+p}{pr}}{r+p}&=-1\\ \frac{-1}{p^2r^2}&=-1\\ r^2&=\frac{1}{p^2}\\ \therefore p=\frac{1}{r}\end{align*}\\ \text{And similarly}\\ s=-\frac{1}{p}$
$\therefore pqrs = p \times -p \times \frac{1}{p}\times -\frac{1}{p}=1\text{ as required.}$
$\text{Note, my diagram was drawn so that }p\text{ and }r\text{ are both positive}\\ \text{Hence }p\neq -r,\text{ rather }p=r\\ \text{If }R\text{ and S}\text{ were }\textit{swapped}\text{, then }r=-\frac{1}{p}$

Holy Crap

Thanks a lot Rui! That was really well explained

armtistic · « **Reply #798 on:** February 04, 2017, 05:59:28 pm »

Hey guys

I know I've been asking a lot of questions here but I have a topic test soon and I want to make sure all my understanding is solid

Could someone help with this conics question?

RuiAce · « **Reply #799 on:** February 04, 2017, 07:27:31 pm »

Quote from: armtistic on February 04, 2017, 05:59:28 pm

Hey guys

I know I've been asking a lot of questions here but I have a topic test soon and I want to make sure all my understanding is solid

Could someone help with this conics question?

Are you sure there hasn't been a typo anywhere?

I attempted this question, and got something catastrophically messy. Not to mention the bad grammar in the first sentence.

Rathin · « **Reply #800 on:** February 05, 2017, 04:11:27 pm »

I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?

Shadowxo · « **Reply #801 on:** February 05, 2017, 05:22:52 pm »

Quote from: Rathin on February 05, 2017, 04:11:27 pm

I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?

I'm not very familiar with those kinds of questions (not in VCE study design) but for an ellipse there is a domain restriction, in this case x can only be between -4 and 4 inclusive or else x² would be more than 16, and the midpoint of the chord cannot be outside the circle, so x must be between -4 and 4
Does this help?

Rathin · « **Reply #802 on:** February 05, 2017, 05:48:29 pm »

Answer says -16/sqrt(17)<=x<=16/sqrt(17) for the restriction..have no idea how they got that.

RuiAce · « **Reply #803 on:** February 05, 2017, 06:07:09 pm »

Quote from: Rathin on February 05, 2017, 04:11:27 pm

I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?

As per this GeoGebra simulation, your locus is a correct one
$\text{Now clearly, the GeoGebra simulation shows a restriction on }c\\ \text{If we move the slider }c\text{ too far to the left}\\ \text{(i.e. too negative)}\\ \text{The line appears to fall outside the ellipse.}\\ \text{This means that we won't have a chord anymore.}\\ \text{Same thing happens the other way.}$
$\text{Return to your working out.}\\ \text{When finding the points of intersection (at the start)}\\ \text{you concluded that}\\ \boxed{17x^2+(16c)x+4(c^2-4)=0}$
$\text{The GeoGebra simulation shows (albeit poorly) that eventually}\\ \text{the line will become a tangent to the ellipse, right before it begins to fall off.}\\ \textbf{And that's the twist}$
$\text{Provided that we clearly have a chord, or a tangent, the midpoint exists.}\\ \text{Without even a tangent, the midpoint does not}\\ \text{Hence, the value of }c\textbf{ must be chosen to ensure that there ARE solutions for }x\\ \text{If there are no solutions for }x\text{ we have no points of intersection and no midpoint.}$
$\text{Hence, the discriminant of the resolving equation should be non-negative.}\\ \begin{align*}(16c)^2-4(17)[4(c^2-4)]&\ge 0\\ 1088-16c^2&\ge 0\\ c^2&\le 68\\ -2\sqrt{17}\le c \le &2\sqrt{17}\end{align*}$
$\text{So since the midpoint is }\left(-\frac{8c}{17},\frac{35c}{17} \right)\\ \text{Noting that }x=-\frac{8c}{17}\text{ we'll play around with the restriction on }c$
$\begin{align*}-2\sqrt{17}&\le -c \le 2\sqrt{17}\\ -\frac{16\sqrt{17}}{17}&\le -\frac{8c}{17} \le\frac{16\sqrt{17}}{17}\\ -\frac{16}{\sqrt{17}} &\le x \le -\frac{16}{\sqrt{17}} \end{align*}\\ \text{Note that ultimately the restriction must be on }x\text{ and/or }y$

Syndicate · « **Reply #804 on:** February 05, 2017, 06:10:11 pm »

Quote from: Rathin on February 05, 2017, 05:48:29 pm

Answer says -16/sqrt(17)<=x<=16/sqrt(17) for the restriction..have no idea how they got that.

Just adding...
For the chord to exist, the line must join two points at the circumference of ellipse, which means it cannot exist at certain points due to its gradient.

RuiAce · « **Reply #805 on:** February 05, 2017, 06:18:04 pm »

Quote from: Syndicate on February 05, 2017, 06:10:11 pm

Just adding...
For the chord to exist, the line must join two points at the circumference of ellipse, which means it cannot exist at certain points due to its gradient.

Pretty sure I addressed this.

Also, it is ok for the two points to coincide.

Quote from: RuiAce on February 05, 2017, 06:07:09 pm

[url=https://ggbm.at/BEZYGzee]
$\text{Provided that we clearly have a chord, or a tangent, the midpoint exists.}\\ \text{Without even a tangent, the midpoint does not}\\ \text{Hence, the value of }c\textbf{ must be chosen to ensure that there ARE solutions for }x\\ \text{If there are no solutions for }x\text{ we have no points of intersection and no midpoint.}$

Syndicate · « **Reply #806 on:** February 05, 2017, 06:25:25 pm »

Quote from: RuiAce on February 05, 2017, 06:18:04 pm

Pretty sure I addressed this.

Also, it is ok for the two points to coincide.

Sorry, I didn't realise you had already posted. I was replying to Shadowxo

QC · « **Reply #807 on:** February 07, 2017, 08:19:49 pm »

Hey, so I was pretty happy with how much of this question I think I got but couldn't get the final result (this was in the last question of 2000 4u HSC). I got the first part then tried t-method, and tried subbing in the hint and equating imaginary parts but my algebra got super messy and I don't exactly know how to get the second part. Is there any faster way using the properties of Imaginary parts etc.? Also, what are your opinions on people not even bothering with these questions cos they would be very tough to get in the exam? My teacher says if you want a safe B6 get all the easy questions (all other than last/8) 100% and look through the rest if you have time. In 4u there is marginal benefit benefit to your atar going from a 90-92 to a 94-95 even though it requires far more effort. My teacher says your better off maximising your 3u or other subjects as there is a huge difference between 90-92 and 94-95 in english, chem, phys, eco etc. Like these questions are fun for me to do so I will probably try them anyway but I can see their point.

RuiAce · « **Reply #808 on:** February 07, 2017, 09:07:10 pm »

Quote from: aroon217 on February 07, 2017, 08:19:49 pm

Hey, so I was pretty happy with how much of this question I think I got but couldn't get the final result (this was in the last question of 2000 4u HSC). I got the first part then tried t-method, and tried subbing in the hint and equating imaginary parts but my algebra got super messy and I don't exactly know how to get the second part. Is there any faster way using the properties of Imaginary parts etc.? Also, what are your opinions on people not even bothering with these questions cos they would be very tough to get in the exam? My teacher says if you want a safe B6 get all the easy questions (all other than last/8) 100% and look through the rest if you have time. In 4u there is marginal benefit benefit to your atar going from a 90-92 to a 94-95 even though it requires far more effort. My teacher says your better off maximising your 3u or other subjects as there is a huge difference between 90-92 and 94-95 in english, chem, phys, eco etc. Like these questions are fun for me to do so I will probably try them anyway but I can see their point.

$\text{It is clear that the LHS falls out immediately}\\ \text{from considering the imaginary part, as if we let}\\ z=\cos \theta + i\sin\theta\\\text{ then by }\textbf{de Moivre's theorem}$
$\begin{align*}\sum_{k=1}^n(z+z^2+\dots+z^k)&=\sum_{k=1}^n\left((\cos\theta+i\sin\theta)+(\cos\theta+i\sin\theta)^2+\dots+(\cos\theta+i\sin\theta)^k\right)\\ &=\sum_{k=1}^n((\cos\theta+i\sin\theta)+(\cos2\theta+i\sin2\theta)+\dots+(\cos k\theta+i\sin k\theta))\\&= \sum_{k=1}^n(\cos\theta+\cos 2\theta + \dots + \cos k\theta)+i\sum_{k=1}^n(\sin \theta+\sin 2\theta+\dots + \sin k\theta)\end{align*}\\ \text{to which the imaginary part is what we required.}\\ \textbf{Question to consider: Why was this clear?}$
$\text{It now falls to find the imaginary part of the RHS.}\\ \text{This requires brute force.}\\ \text{Also note that squaring the given result gives us}\\ \boxed{\frac{z^2}{(1-z)^2}=\frac{-1}{4\sin^2 \frac{\theta}{2}}(\cos \theta+i\sin \theta)}\\ \text{having used de Moivre's theorem again.}$
$\text{Remember that we only want the imaginary part}\\ \text{Hence, we attempt to make explicit anything with }i\text{ in it.}\\\begin{align*}&\quad\frac{nz}{1-z}-\frac{z^2}{(1-z)^2}(1-z^n)\\&= \frac{ni}{2\sin \frac{\theta}{2}}\left(\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}\right)+\frac{1}{4\sin^2 \frac{\theta}{2}}(\cos \theta+i\sin \theta)(1-\cos n\theta - i \sin n\theta)\\ &=\frac{n}{2\sin \frac{\theta}{2}}\left(i\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)+\frac{\cos \theta(1-\cos n\theta)+i\sin \theta(-i\sin n\theta)}{4\sin^2 \frac{\theta}{2}}+\frac{i\sin\theta (1-\cos n\theta)+\cos \theta(-i\sin n\theta)}{4\sin^2 \frac{\theta}{2}}\end{align*}$
$\text{So taking all we need}\\ \text{Im}\left(\frac{nz}{1-z}-\frac{z^2}{(1-z)^2}(1-z^n)\right)=\frac{n\cos \frac{\theta}{2}}{2\sin \frac{\theta}{2}}+\frac{\sin \theta(1-\cos n\theta)-\cos \theta \sin n\theta}{4\sin^2\frac{\theta}2}\\ \text{Some work involving trigonometric identities will be used in tidying this up.}$
$\begin{align*}&\quad \text{Im}\left(\frac{nz}{1-z}-\frac{z^2}{(1-z)^2}(1-z^n)\right)\\ &= \frac{2n\cos \frac{\theta}{2}\sin \frac{\theta}2}{4\sin^2 \frac{\theta}{2}}+\frac{\sin \theta - (\sin \theta \cos n\theta + \cos \theta \sin n\theta)}{4\sin^2\frac{\theta}{2}} \\ &=\frac{n \sin \theta}{4\sin^2\frac{\theta}{2}}+\frac{\sin \theta - \sin (n+1)\theta}{4\sin^2 \frac{\theta}{2}}\\ &= \frac{(n+1)\sin \theta + \sin (n+1)\theta}{4\sin^2\frac{\theta}2} \end{align*}\\ \text{Equate back with the LHS and we are done.}$
Maybe a few strategies with de Moivre's theorem, but otherwise what strategy. I just brute forced it and hoped for the best.

Also, yeah I agree with your teacher. Especially since this was before 2001 (even if it was barely before 2001) and no longer reflects the difficulty of the HSC. I only did the questions up to the new version of the HSC.

QC · « **Reply #809 on:** February 07, 2017, 09:26:48 pm »

oh dang I didn't apply de Moivres again with the (costheta+isintheta)^2 only the power n so i was getting random half powers and I couldn't get it. Thanks man, its pretty annoying this stuff isnt online

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