HSC 2012 MX2 Marathon (archive) (2 Viewers)

IamBread · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

m<0 only applies for 0 < x < a which isn't a condition, so why is m<0?

math man · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

The equation of the ellipse is the top semi-ellipse
Covering the first two quadrants. Now when you sub
Y=mx+c into semi ellipse and form a quadratic in x
The y int of the tangent must be positive otherwise
It can't be a tangent to the top semi ellipse so c>0, now
As for m<0 there is only one possible way this can
Occur and that is if the x coordinate of the point of
Contact of the tangent is positive. This is deduced if
You find the x value of the point of contact using
The double root method, and this shows the x value of the
Poiny of contact is positive if m<0 otherwise the x value
Will be negative, so this condition for m can't occur unless
The tangent interests in the first quadrant

math man · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Spirals question is flawed for the m condition
The way it is worded

math man · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

derivative is taken to apply double root method for solving to find x value of point of contact.

Nooblet94 · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

MOAR CONICS!!

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $(Q1)$\\ $Prove that the pair of tangents from the point $ P(4,5)$ to the ellipse $ \frac{x^2}{25}@plus;\frac{y^2}{16}=1 $ are at right angles to one another$~\\ \\ $(Q2)$\\ $Obtain the quadratic equation satisfied by $m$ where $m$ is the gradient of the tangent $ y=mx@plus;c$ from the external point $P(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}@plus;\frac{y^2}{b^2}=1 $. Hence, find the locus of $P$ if the two tangents from $P$ are at right angles.$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $(Q1)$\\ $Prove that the pair of tangents from the point $ P(4,5)$ to the ellipse $ \frac{x^2}{25}+\frac{y^2}{16}=1 $ are at right angles to one another$~\\ \\ $(Q2)$\\ $Obtain the quadratic equation satisfied by $m$ where $m$ is the gradient of the tangent $ y=mx+c$ from the external point $P(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $. Hence, find the locus of $P$ if the two tangents from $P$ are at right angles.$" title="\\ $(Q1)$\\ $Prove that the pair of tangents from the point $ P(4,5)$ to the ellipse $ \frac{x^2}{25}+\frac{y^2}{16}=1 $ are at right angles to one another$~\\ \\ $(Q2)$\\ $Obtain the quadratic equation satisfied by $m$ where $m$ is the gradient of the tangent $ y=mx+c$ from the external point $P(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $. Hence, find the locus of $P$ if the two tangents from $P$ are at right angles.$" /></a>

nightweaver066 · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
MOAR CONICS!!

$\\ $(Q1)$\\ $Prove that the pair of tangents from the point $ P(4,5)$ to the ellipse $ \frac{x^2}{25}+\frac{y^2}{16}=1 $ are at right angles to one another$~\\ \\ $(Q2)$\\ $Obtain the quadratic equation satisfied by $m$ where $m$ is the gradient of the tangent $ y=mx+c$ from the external point $P(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $. Hence, find the locus of $P$ if the two tangents from $P$ are at right angles.$$

$1) E: \frac{x^2}{25} + \frac{y^2}{16} = 1$

$Let $ L: y = mx + c $ be a tangent to E$

$$As L passes through (4, 5), $ c = 5 - 4m$

$Solving L and E for intercepts,$

$\frac{x^2}{25} + \frac{(mx + c)^2}{16} = 1$

$16x^2 + 25(m^2x^2 + 2mcx + c^2) = 400$

$(16 + 25m^2)x^2 + 50mcx + 25c^2 - 400 = 0$

$\Delta = 2500m^2c^2 - 4(16 + 25m^2)(25c^2 - 400)$

$$For L to be a tangent, $ \Delta = 0$

$25m^2c^2 - (16 + 25m^2)(c^2 - 16) = 0$

$25m^2c^2 - 16c^2 + 256 - 25m^2c^2 + 400m^2 = 0$

$400m^2 - 16(5 - 4m)^2 + 256 = 0$

$100m^2 - 4(25 - 40m + 16m^2) + 64 = 0$

$100m^2 - 100 - 160m - 64m^2 + 64 = 0$

$36m^2 - 160 - 36 = 0$

$m = m_1 \; , \; m_2$

$m_1\times m_2 = \frac{-36}{36} = -1$

$\therefore $ The tangents are at right angles to one another$$

$2) E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$Let L: y = mx + c be a tangent to E$

$$As L passes through $ P(x_1, y_1),$

$c = y_1 - mx_1$

$Solving E and L for intercepts$

$b^2x^2 + (mx + c)^2a^2 = a^2b^2$

$b^2x^2 + (m^2x^2 + 2mcx + c^2)a^2 = a^2b^2$

$b^2x^2 + a^2m^2x^2 + 2a^2mcx + a^2c^2 - a^2b^2 = 0$

$\Delta = 4a^4m^2c^2 - 4(b^2 + a^2m^2)(a^2c^2 - a^2b^2)$

$$For L to be tangent, $ \Delta = 0$

$a^2m^2c^2 - (b^2 + a^2m^2)(c^2 - b^2)= 0$

$a^2m^2c^2 - b^2c^2 + b^4 - a^2m^2c^2 + a^2m^2b^2 = 0$

$a^2m^2 - c^2 + b^2 = 0$

$a^2m^2 + b^2 - (y_1 - mx_1)^2 = 0$

$a^2m^2 + b^2 - y_1^2 + 2x_1y_1m - m^2x_1^2 = 0$

$(a^2 - x_1^2)m^2 + 2x_1y_1m + b^2 - y_1^2 = 0$

$$If the two tangents from P are at right angles, $ m_1 \times m_2 = -1$

$m_1 \times m_2 = \frac{b^2 - y_1^2}{a^2 - x_1^2} = -1$

$b^2 - y_1^2 = x_1^2 - a^2$

$x_1^2 + y_1^2 = a^2 + b^2$

$As $ x_1 \; \& \; y_1 $ are variable points, the locus of P is the director's circle, i.e.$

$x^2 + y^2 = a^2 + b^2$

OMGITzJustin · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

what is the best method of graphing 1/f(x) graphs? for example 1/sinx?

IamBread · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

First graph f(x) then using that graph y=1/f(x)

math man · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

OMGITzJustin said:
what is the best method of graphing 1/f(x) graphs? for example 1/sinx?

f(x) represents the y values and 1/f(x) means 1 over every y value. So you use limits as f(x) appraoches 0 and infinity and
note when f(x) equals 0 there becomes vertical asymptotes. Also the turning points on f(x) stay the same, only the y value changes
and the type of turning point reverses

Trebla · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
$$Show that for $0 < x < \frac{\pi}{2}\\\\\displaystyle\int \dfrac{1+\sin x}{\cos x}\,dx = \displaystyle\sum_{n=1}^{\infty}\dfrac{\sin^nx}{n}+c$

.

IamBread · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

I thought I posted by solution for that, guess I'll do it now

.

$\int\frac{1+sinx}{cos}dx \\\\ let u = sinx \\\\ \int\frac{1+sinx}{cos}dx = \int \frac{1+u}{1-u^2}du \\\\ = \int\frac{1}{1-u}du \\\\ = -ln|1-u| + c \\\\ =-ln|1-sinx|+c$

$Using $ lnx = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n, \\\\ -ln|1-sinx| = \sum_{n=1}^\infty \frac{(-1)^n}{n}(-sinx)^n \\\\ =\sum_{n=1}^\infty \frac{(-1)^n}{n}(-1)^n(sinx)^n \\\\ = \sum_{n=1}^\infty \frac{sin^n x}{n} $ as required.$

Trebla · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Try to do it using only HSC content. The solution arrives quicker though it might be trickier to figure out.

deterministic · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

question kinda hints to use GP

Noting that $|\sin x|< 1$ for $0 < x < \frac{\pi}{2}$ , see that:
$\int \frac{\sin x+1}{\cos x} dx &= \int \frac{1-\sin^2 x}{\cos x (1-\sin x)} dx = \int \frac{\cos x}{1-\sin x} dx=\int \cos x \sum_{n=0}^\infty \sin^n x dx$

Assuming interchangeability between integral and summation, the result follows easily.

cutemouse · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Trebla likes his GP type of questions

largarithmic · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Trebla's original question with limits 0 --> pi/2 was actually convergent. Just because a point is undefined, does not necessarily mean that the area is automatically infinite.

No its not...

$\int_0^d \frac{1+\sin x}{\cos x} dx = \int_0^d (\tan x + \sec x) dx$
$= \left[-\ln |\cos x| + \ln |\tan x + \sec x| \right]_0^d = \left[ \ln \left(\frac{\tan x + \sec x}{\cos x} \right)\right]_0^d$
$= \ln \left|\frac{\tan d + \sec d}{\cos d}\right|$

Then as d approaches pi/2, cosd -> 0, and tand + secd -> positive infinity. So its pretty clear that the integral has to diverge

math man · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
No its not...

$\int_0^d \frac{1+\sin x}{\cos x} dx = \int_0^d (\tan x + \sec x) dx$
$= \left[-\ln |\cos x| + \ln |\tan x + \sec x| \right]_0^d = \left[ \ln \left(\frac{\tan x + \sec x}{\cos x} \right)\right]_0^d$
$= \ln \left|\frac{\tan d + \sec d}{\cos d}\right|$

Then as d approaches pi/2, cosd -> 0, and tand + secd -> positive infinity. So its pretty clear that the integral has to diverge

This does seem true, but e ing both
Sides then applying l hopitals rule
Will show this is actually convergent
As carrot said

seanieg89 · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Umm no, largarithmic is correct...the integral does not converge. The hypotheses of L'Hospitals rule are not even satisfied here.

Carrotsticks · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

Beforehand I was being lazy and checked the convergence using Wolfram (which claimed it converges and actually gave a value) instead of actually verifying it by hand.

Never again will I trust it for anything more than simple calculations ^^

AAEldar · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

I don't want to spoil the party but...

Any chance we could have questions more orientated towards a regular MX2 student, or a little harder, but something that most should at least be able to attempt. A lot of people would probably be scared off, especially since you guys are into or out of uni maths already!

Carrotsticks · Feb 26, 2012

Re: 2012 HSC MX2 Marathon

AAEldar said:
I don't want to spoil the party but...

Any chance we could have questions more orientated towards a regular MX2 student, or a little harder, but something that most should at least be able to attempt. A lot of people would probably be scared off, especially since you guys are into or out of uni maths already!

Something I predict will appear in either the school Trials or the Complex Number exam term test (verified appeared in already 3 schools as per my prediction last year haha):

Find the value of:

$i^{2012}$

OR

Find the value of:

$\sum_{k=0}^{2012}i^{k}$

Even better:

$\prod_{k=0}^{2012}i^{k}$

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