HSC 2012 MX2 Marathon (archive) (1 Viewer)

math man · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

The skill to using the reverse chain rule
is to be very good at recognising powers of
functions and their derivatives.

SpiralFlex · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Some Conics...

$\alpha)\; $Let$\; a\;$ and $\;b \;$ be positive constants.$$

$$Consider the curve E:$ \; y=\frac{b}{a} \sqrt{a^2-x^2}. \; $Prove that the straight line$ \; y=mx+c $\; is a tangent to$ \;E \;$if and only if$ \; m<0 \; $and$ \; c=\sqrt{a^2m^2+b^2}\; \; \; \fbox{4}$

$\beta)\; $Consider the curve$ \; E_{1}: y=\sqrt{27-3x^2}, \; $where$ \; 0<x<3,\; $and the curve$\; E_{2}:y=\sqrt{9-\frac{x^2}{3}}\; $where$ \;0<x<3\sqrt{3}$

$$a) Find the point of intersection of \; $ \;E_{1} \; $and$ \; E_{2}\; \; \; \fbox{2}$

$$b) Let$ \; L\; $be the common tangent to $ \;E_{1} \; $and$ \; E_{2}\; \; \;$ .

$$\;i) Find the equation of$\; L.\; \; \;\fbox{2}$

$$\; ii) Find the area of the region bounded by $\;E_{1} \;, \; E_{2} \; $and$ \; L.\; \; \; \fbox{2}$

Nooblet94 said:
For part alpha I can't work out why m<0. I can see it must be the case for the rest of the question because of the domain.

You can do it.

Other attempts pl0x. I like this question.

Nooblet94 · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
You can do it.

Other attempts pl0x. I like this question.

I enjoyed the question too - I did the rest of it last night after skipping the first part (which I still can't get

)

SpiralFlex · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Use the force.

Carrotsticks · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
I did the rest of it last night after skipping the first part (which I still can't get )

Substitute y=mx+c into the equation of the semi-ellipse and make a quadratic in terms of x.

Let the discriminant be equal to zero and after a whole lot of algebra, you should get the solution.

However, I do not understand why m < 0 is a condition. I do not see why positive gradients are excluded as possible tangents.

This is only the case if we purely consider the first quadrant of the Ellipse in canonical form and the equation you provided covers the first 2 quadrants.

Nooblet94 · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Substitute y=mx+c into the equation of the semi-ellipse and make a quadratic in terms of x.

Let the discriminant be equal to zero and after a whole lot of algebra, you should get the solution.

However, I do not understand why m < 0 is a condition. I do not see why positive gradients are excluded as possible tangents.

This is only the case if we purely consider the first quadrant of the Ellipse in canonical form and the equation you provided covers the first 2 quadrants.

That's exactly what I was struggling with (why m<0), I got the rest of it, I just assumed I was missing something since Spiral's almost always correct and nobody else had mentioned anything.

IamBread · Feb 25, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
For part alpha I can't work out why m<0.

Lol I'm stuck at this point too, can't work that out.

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.

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