Author Topic: 4U Maths Question Thread (Read 665366 times) Tweet Share

RuiAce · « **Reply #1725 on:** February 23, 2018, 07:55:47 pm »

Quote from: arii on February 23, 2018, 05:12:37 pm

Not sure how to continue here...

Normally I'd try to factorise out sqrt{k+1} but it seems to complicate it even more...

$\text{Unfortunately there really isn't any slick way of doing this one.}\\ \text{I just used working backwards to brute force it.}$

Working backwards

$\text{Essentially, our ultimate goal is to prove that}\\ \frac{4k+3}{6}\sqrt{k} + \sqrt{k+1} \le \frac{4k+7}{3}\sqrt{k+1}$
$\text{To work backwards, suppose this is true. Then,}\\ \begin{align*}(4k+3)\sqrt{k} + 6\sqrt{k+3} &\le (4k+7)\sqrt{k+1}\\ (4k+3)\sqrt{k} &\le (4k+1)\sqrt{k+1}\\ (16k^2+12k+9)k&\le (16k^2+8k+1)(k+1)\\ 16k^3+12k^2+9k &\le 16k^3 + 24k^2 + 9k + 1\\ 0&\le 12k^2 + 1\end{align*}$
$\text{Well, that's nice, because }12k^2+1\ge 0\text{ is }\textbf{obviously}\text{ true.}\\ \text{So now let's write it all backwards.}$

$\textit{Proof:}\text{ Since for all real }k, \text{ we know that }12k^2+1\ge 0\text{, we then have}\\ \begin{align*}0&\le 12k^2+1\\ 16k^3+12k^2+9k &\le 16k^3+24k^2+9k+1\\ k(4k+3)^2 &\le (k+1)(4k+1)^2\\ (4k+3)\sqrt{k}&\le (4k+1)\sqrt{k+1}\end{align*}\\ \text{having taken positive square roots since all quantities are positive.}$
$\text{Now, consider}\\ \begin{align*}LHS &= \sqrt{1}+\dots + \sqrt{k}+\sqrt{k+1}\\ &\le \frac{(4k+3)\sqrt{k}}{6} + \sqrt{k+1}\tag{inductive hypothesis}\\ &\le \frac{(4k+1)\sqrt{k+1}}{6} + \sqrt{k+1}\\ &= \frac{(4k+7)\sqrt{k+1}}{6}\\ &= RHS\end{align*}$

justwannawish · « **Reply #1726 on:** February 23, 2018, 10:06:30 pm »

Hey,

any general types on drawing y=ln(e^f(x)) where a graph of f(x) is provided for you?

also drawing e^ln(f(x))

RuiAce · « **Reply #1727 on:** February 23, 2018, 10:12:06 pm »

Quote from: justwannawish on February 23, 2018, 10:06:30 pm

Hey,

any general types on drawing y=ln(e^f(x)) where a graph of f(x) is provided for you?

also drawing e^ln(f(x))

That's literally just going to be $ y = f(x) $ in both cases.

(Although the latter has an extra restriction - $ \ln (e^\alpha)= \alpha$ always, but $e^{\ln \alpha} = \alpha$ only when $\alpha > 0$, just because of the domain of $\ln x$ being $ x > 0$)

amaher · « **Reply #1728 on:** February 24, 2018, 02:44:24 pm »

Hi!!! I'm struggling with a conics q about ellipses -
M is the midpoint of PQ where P and Q lie on x²/a² + y²/b² = 1. O is the centre of the ellipse. Show that the tangents at P and Q intersect on OM produced.
Thank you to anyone that helps!!

RuiAce · « **Reply #1729 on:** February 24, 2018, 04:23:51 pm »

Quote from: amaher on February 24, 2018, 02:44:24 pm

Hi!!! I'm struggling with a conics q about ellipses -
M is the midpoint of PQ where P and Q lie on x²/a² + y²/b² = 1. O is the centre of the ellipse. Show that the tangents at P and Q intersect on OM produced.
Thank you to anyone that helps!!

$\text{Assume that }P\text{ is }(a\cos \theta, b\sin \theta)\text{ and }Q\text{ is }(a\cos \phi, b\sin \phi)\\ \text{so that }M\text{ is the point }\left(\frac{a}{2}(\cos\theta + \cos\phi), \frac{b}{2}(\sin\theta+\sin\phi) \right)$
$\text{It is then easy to deduce that }OM\text{ has equation}\\ \boxed{y = \frac{b(\sin\theta+\sin\phi)}{a(\cos\theta+\cos\phi)}x}$
_________________________________________________________
$\text{Now, the equation of the tangents at }P\text{ and }Q\text{ respectively are}\\ \begin{align*}\frac{x\cos \theta}{a}+\frac{y\sin\theta}{b}\tag{1}&=1\\ \frac{x\cos\phi}{a}+\frac{y\sin\phi}{b}&=1\tag{2}\end{align*}\\ \text{which should be easy to prove.}$
$\text{The usual approach here would be to find the point of intersection}\\ \text{via simultaneous equations (most likely substitution)}\\ \text{and then show that it satisfies the equation of }OM.\\ \textit{However}\text{, I will go for a slicker method.}$
_________________________________________________________

Note that the intuition to do this was via working backwards. Here, the details in working backwards will be omitted.
$\text{Firstly, starting from the obvious statement }\boxed{1=1}\text{ we deduce that}\\ \begin{align*}\cos^2\theta+\sin^2\theta &= \cos^2\phi + \sin^2\phi\\ \sin^2\theta - \sin^2\phi &= -\cos^2\theta + \cos^2 \phi\\ (\sin\theta-\sin\phi)(\sin\theta+\sin\phi)&=-(\cos\theta+\cos\phi)(\cos\theta-\cos\phi)\\ \therefore \frac{\sin\theta+\sin\phi}{\cos\theta+\cos\phi}&=-\frac{\cos\theta-\cos\phi}{\sin\theta-\sin\phi}\tag{3}\end{align*}$
$\text{The point of intersection must satisfy both }(1)\text{ and }(2).\\ \text{Therefore, it also satisfies }(1)-(2)\\ \text{i.e. }\boxed{\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}-\frac{x\cos\phi}{a}-\frac{y\sin\phi}{b}=0}$
$\text{It turns out, that by rearranging this, we can show that}\\ \text{the coordinates of the points of intersection}\\ \text{actually satisfy the equation of }OM\text{, which is what we require:}\\ \begin{align*}\frac{x}{a}(\cos\theta-\cos\phi)+\frac{y}{b}(\sin\theta-\sin\phi) &= 0\\ \frac{y}{b}(\sin\theta-\sin\phi)&= -\frac{x}{a}(\cos\theta-\cos\phi)\\ y&= -\frac{b(\cos\theta-\cos\phi)}{a(\sin\theta-\sin\phi)} x\\ &=-\frac{b(\sin\theta+\sin\phi)}{a(\cos\theta+\cos\phi)}x\end{align*}$
$\text{So since the equation of }OM\text{ is satisfied,}\\ \text{the point of intersection must lie on }OM\text{ produced.}$

kaustubh.patel · « **Reply #1730 on:** February 26, 2018, 01:00:09 am »

Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.

RuiAce · « **Reply #1731 on:** February 26, 2018, 07:31:28 am »

Quote from: kaustubh.patel on February 26, 2018, 01:00:09 am

Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.

I'll come back to this with a proper answer in a few hours time. Basically the idea is that you start by just finding the equation of the chord of contact from (0,4), then locating the points of intersection between the chord of contact and the hyperbola

Edit:
\[\text{The usual method to find the chord of contact}\\ \text{can be found in the previous lecture slides.}\]
(Or alternatively in the book.)
$\text{Of course, seeing as though we have a hyperbola, it will take the form}\\ \boxed{\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1}\\ \text{In our case, we have }a=3, b=6\text{ and }(x_0,y_0) = (0,4)\\ \text{and thus the chord of contact is }\boxed{-\frac{y}{9} = 1}\\ \text{i.e. }\boxed{y=-9}$
Note: The values of $a$ and $b$ were obtained because we can only infer them when the RHS of the equation is 1. So the equation of the hyperbola had to be rearranged into $ \frac{x^2}{9} - \frac{y^2}{36} = 1$
$\text{From here, the question is easy, because we just sub }y=-9\text{ in:}\\ \frac{x^2}{9} - \frac{81}{36} = 1 \implies \boxed{x = \pm \frac{3\sqrt{13}}{2} }$
And thus our points are $ \left(\pm\frac{3\sqrt{13}}2,-9 \right) $

jazzycab · « **Reply #1732 on:** February 26, 2018, 11:20:54 am »

Quote from: kaustubh.patel on February 26, 2018, 01:00:09 am

Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.

Let's say that the solution points are at $\left(\alpha,\beta\right)\ \left(1\right)$. From the equation of the curve, we get:
$4\alpha^2-\beta^2=36\\\beta^2=4\left(\alpha^2-9\right)\\\beta=\pm2\sqrt{\alpha^2-9}\ \left(2\right)$
Secondly, we can find the gradients of the tangents by implicitly differentiating the hyperbola:
$\frac{d}{dx}\left(4x^2\right)-\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(36\right)\\8x-2y\frac{dy}{dx}=0\\\frac{dy}{dx}=\frac{4x}{y}$
Substituting in $\left(1\right)$ and $\left(2\right)$ gives the gradients:
$m=\frac{4\alpha}{\pm2\sqrt{\alpha^2-9}}\\=\pm\frac{2\alpha}{\sqrt{\alpha^2-9}}\ \left(3\right)$
The equation of each tangent line is given by $y_T=m\left(x-\alpha\right)+\beta$. We can subsequently substitute in $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$:
$y_T=\pm\frac{2\alpha}{\sqrt{\alpha^2-9}}\left(x-\alpha\right)\pm2\sqrt{\alpha^2-9}\ \left(4\right)$
We know that the tangent lines pass through $\left(0,4\right)$:
$4=\pm\frac{2\alpha}{\sqrt{\alpha^2-9}}\left(-\alpha\right)\pm2\sqrt{\alpha^2-9}\\4\sqrt{\alpha^2-9}=\mp2\alpha^2\pm2\left(\alpha^2-9\right)\\4\sqrt{\alpha^2-9}=\mp18\\4\sqrt{\alpha^2-9}>0\\\Rightarrow4\sqrt{\alpha^2-9}=18\\\sqrt{\alpha^2-9}=\frac{9}{2}\\\alpha^2-9=\frac{81}{4}\ \left(5\right)\\\alpha^2=\frac{117}{5}\\\alpha=\pm\frac{3\sqrt{13}}{2}\ \left(6\right)$
Substituting $\left(5\right)$ into $\left(1\right)$ gives:
$\beta=\pm2\sqrt{\frac{81}{4}}\\=\pm2\times\frac{9}{2}=\pm9\ \left(7\right)$
Given that I have squared and square rooted both sides of equations a significant number of times, we will need to check what combinations of $\left(\alpha,\beta\right)$ are correct. Let's look at the tangent line equation, $\left(4\right)$, which passes through $\left(0,4\right)$:
$\text{Let }\alpha=\frac{3\sqrt{13}}{2}\\y_T=\frac{4\times\frac{3\sqrt{13}}{2}}{\beta}\left(x-\frac{3\sqrt{13}}{2}\right)+\beta\\4=\frac{4\times\frac{3\sqrt{13}}{2}}{\beta}\left(-\frac{3\sqrt{13}}{2}\right)+\beta\\4=-\frac{117}{\beta}+\beta\ \left(8\right)\\4\beta=-117+\beta^2\\\beta^2-4\beta-117=0\\\left(\beta-13\right)\left(\beta+9\right)=0\\\Rightarrow\beta=-9\text{ or }13$
From $\left(7\right)$ we can see that $\alpha=\frac{3\sqrt{13}}{2}\Rightarrow\beta=-9$. Now let's investigate for the other $x$-value:
$\text{Let }\alpha=-\frac{3\sqrt{13}}{2}\\y_T=\frac{-4\times\frac{3\sqrt{13}}{2}}{\beta}\left(x-\left(-\frac{3\sqrt{13}}{2}\right)\right)+\beta\\4=\frac{-4\times\frac{3\sqrt{13}}{2}}{\beta}\left(\frac{3\sqrt{13}}{2}\right)+\beta\\4=-\frac{117}{\beta}+\beta\ \left(9\right)$
Note that $\left(9\right)$ is identical to $\left(8\right)$ so it has the same solutions. Thus, $\alpha=-\frac{3\sqrt{13}}{2}\Rightarrow\beta=-9$. Hence, the two points on the hyperbola whose tangents intersect at $\left(0,4\right)$ are:
$\left(\frac{3\sqrt{13}}{2},-9\right)\text{ and }\left(-\frac{3\sqrt{13}}{2},-9\right)$

kaustubh.patel · « **Reply #1733 on:** February 26, 2018, 08:08:18 pm »

Holy crap me, Jassycab, you broke it down so wonderfully, my only fear is that in an exam i may use the chord of contact method because its a bit simpler but forming a tangent and subbing in the values is pretty darn insightful.

jazzycab · « **Reply #1734 on:** February 26, 2018, 09:29:52 pm »

Quote from: kaustubh.patel on February 26, 2018, 08:08:18 pm

Holy crap me, Jassycab, you broke it down so wonderfully, my only fear is that in an exam i may use the chord of contact method because its a bit simpler but forming a tangent and subbing in the values is pretty darn insightful.

We don't look at the chord of contact method in the VCE, so I just used what I could from the VCE Mathematics syallbi

kaustubh.patel · « **Reply #1735 on:** February 27, 2018, 09:34:06 am »

Hey maths lovers another conics question if you guys have time.

RuiAce · « **Reply #1736 on:** February 27, 2018, 10:52:13 am »

Quote from: kaustubh.patel on February 27, 2018, 09:34:06 am

Hey maths lovers another conics question if you guys have time.

$\text{The first things to infer are that the foci are at}\\ 1+3i\text{ and }9+3i$
i.e. the points $ (1,3) $ and $ (9,3) $.
$\text{Additionally, }2a = 10\text{, so }\boxed{a=5}.$
_______________________________________
$\text{The centre of the ellipse is just the midpoint of the two foci.}\\ \text{This will be }\boxed{(5,3)}\\ \text{and hence the complex number we require is }\boxed{5+3i}$
_______________________________________
$\text{Since the distance from the ellipse's centre to any foci is }4\\ \text{we now know that }\boxed{ae = 4}.\\ \text{Recalling that }a = 5\text{, we see that }\boxed{e = \frac{4}{5}}$
$\text{Hence, via }\boxed{b^2 = a^2(1-e^2)}\\ \text{we see that }\boxed{b = 3}.$
$\text{Therefore our ellipse has equation}\\ \boxed{ \frac{(x-5)^2}{25} + \frac{(y-3)^2}9 = 1}$
The length of the major axis is 10, and the length of the minor axis is 6.

_______________________________________
$\text{Now, recall that }\arg z\text{ measures the }\textbf{angle}\\ \text{the complex number }z\text{ makes, at }0+0i\text{, with respect to the positive real axis.}$
$\text{Effectively speaking, the minimum and maximum values of }\arg z\textbf{ will always occur}\\ \text{on the }\textbf{boundary}\text{ of the shape we have.}$
For the ellipse, the boundary is pretty much just going to be the ellipse itself.
$\text{Additionally, the min and max values are attained}\\ \text{when the ray drawn from the origin}\\ \textbf{is tangential to the ellipse.}$
$\text{From the diagram, it is clear that the tangents to the ellipse, from }0+0i,\\ \text{are actually just }y=0\text{ and }x=0.$
_______________________________________
$\text{Points on the line }y=0\text{ will satisfy }\arg z = 0.$
(Note: We only care about the right-side of the x-axis. The reason we don't care about the left side is simply because there just is no ellipse there.)
$\text{Points on the line }x=0\text{ will satisfy }\arg z = \frac\pi2.$
$\text{Therefore, the range of values for }\arg z\\ \text{will be }0\le \arg z \le \frac\pi2$
More examples of types of problems like part iii)

kaustubh.patel · « **Reply #1737 on:** February 28, 2018, 11:32:56 pm »

Oh sweet the arg part i can see clearly it's angle made by the x and y axis to the +ve x-axis, the diagram makes it much clear. I'd like to request help for 2 more questions though, again conics.

kaustubh.patel · « **Reply #1738 on:** March 01, 2018, 12:01:40 am »

Oh for question 12 i finally got a sorta solution please advise me if it's a preferable method.

first I prove a^2m^2+b^2=k^2, k is the y intercept of the tangent, (in the attached image)
and simply sub in a= 5, b=4 and m=1 and solve for k to put in eq. y=x+k

RuiAce · « **Reply #1739 on:** March 01, 2018, 12:49:52 am »

Quote from: kaustubh.patel on February 28, 2018, 11:32:56 pm

Oh sweet the arg part i can see clearly it's angle made by the x and y axis to the +ve x-axis, the diagram makes it much clear. I'd like to request help for 2 more questions though, again conics.

Just from memory I'm pretty sure you arrived at the best method for Q12. (I remember seeing that $a^2m^2+b^2=k^2$ thing, or something very similar to it quite a lot.)
$\text{The midpoint satisfies}\\ \begin{align*}x &=\frac{a\sec\theta+a\sec\phi}{2}\\ y&=\frac{b\tan\theta+b\tan\phi}{2} \end{align*}$
$\text{Which we will rearrange into}\\ \begin{align*}\frac{x}{a} &= \frac{1}{2}(\sec\theta + \sec \phi)\\ \frac{y}{b}&= \frac{1}{2} (\tan\theta+\tan\phi) \end{align*}\\ \text{Now, keeping in mind that }\boxed{\theta+\phi = \frac\pi2}\text{, we do battle with algebra.}$
_____________________________________________________
\begin{align*}&\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}\\ &= \frac14 \left[(\sec\theta+\sec\phi)^2 - (\tan\theta + \tan\phi)^2 \right]\\ &= \frac14 \big[ (\sec^2\theta-\tan^2\theta) + (\sec^2\phi-\tan^2\phi) \\ &\quad \quad+ 2(\sec\theta\sec\phi - \tan\theta\tan\phi)\big]\\ &= \frac12 \left[ 1 +\sec\theta\sec\phi-\tan\theta\tan\phi \right]\end{align*}
\begin{align*}&\therefore \frac{x^2}{a^2}-\frac{y^2}{b^2} - \frac{y}{b}\\ &= \frac12 (1+\sec\theta\sec\phi-\tan\theta\tan\phi-\tan\theta-\tan\phi)\\ &= \frac12 \left( \frac{\cos\theta\cos\phi+1-\sin\theta\sin\phi-\sin\theta\cos\phi-\cos\theta\sin\phi}{\cos\theta\cos\phi} \right)\\ &= \frac12 \left(\frac{\cos(\theta+\phi) +1- \sin(\theta+\phi)}{\cos\theta\cos\phi}\right)\\ &=\frac12 \left(\frac{\cos \frac\pi2 + 1 - \sin \frac\pi2}{\cos\theta\cos\phi} \right) \\ &= \frac12 \left( \frac{0 + 1 - 1}{\cos\theta\cos\phi} \right)\\ &= 0\end{align*}
\[ \therefore \boxed{\frac{x^2}{a^2}-\frac{y^2}{b^2} = \frac{y}{b}}\text{ as required.} \]
Note: You are not expected to solve locus problems for the ordinary hyperbola (and ellipse) for the HSC.

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