HSC 2015 MX2 Marathon (archive) (3 Viewers)

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION (EASIER):

$If $|z+2|+|z-2|=8$ find the equation of the locus of the point $P$ representing $z$ on an Argand diagram.$

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$

$\frac{2x}{a^2} + \frac{2y \times y'}{b^2}=0$

$\therefore y'= \frac{-2b^2x_1}{2y_1a^2}$

$$ Focal chord gradient = $ \frac{y_1}{x_1-ae}$

$Use equations $ b^2=a^2(1-e^2) $ to equate$

Yeah, better you try this at home with a clear mind, draw a diagram and try to see what the question actually means, I am not sure if your method of equating gradients ?? is actually correct, since the question at hand refers to angles other than that from the positive x-axis..

Ekman · Feb 13, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION (EASIER):

$If $|z+2|+|z-2|=8$ find the equation of the locus of the point $P$ representing $z$ on an Argand diagram.$

$x=\pm2 $ is the focus points on the locus, and their addition is equal to 8$$

$\therefore a=4 $ and $ e=\frac{1}{2}$

$\therefore b^2 = 12$

$\frac{x^2}{16} + \frac{y^2}{12} = 1$

Sorry for being vague in my working out...

FrankXie · Feb 13, 2015

Re: HSC 2015 4U Marathon

$The line $ y=mx+b $ is tangential to the ellipse $ \frac{x^2}{9}+\frac{y^2}{4}=1. $ Show that $ b^2=9m^2+4. $ Hence find the minimum area bounded by the tangent and the coordinate axes.$

Carrotsticks · Feb 13, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
$The line $ y=mx+b $ is tangential to the ellipse $ \frac{x^2}{9}+\frac{y^2}{4}=1. $ Show that $ b^2=9m^2+4. $ Hence find the minimum area bounded by the tangent and the coordinate axes.$

Surely doing this parametrically is faster, but I guess for the sake of trying a different technique it's good practise.

Ekman · Feb 13, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
$The line $ y=mx+b $ is tangential to the ellipse $ \frac{x^2}{9}+\frac{y^2}{4}=1. $ Show that $ b^2=9m^2+4. $ Hence find the minimum area bounded by the tangent and the coordinate axes.$

For the first part sub in y=mx+b into the ellipse equation and make discriminant = 0:

$4x^2 + 9y^2=36$

$x^2(4+9m^2) + 18mbx+9b^2-36 =0$

$324m^2b^2-4(4+9m^2)(9b^2-36)=0$

$\therefore b^2=9m^2+4$

$y=b $ and $ x=\frac{-b}{m}$

$\therefore Area=\frac{-b^2}{2m}$

$=\frac{-9m^2-4}{2m}$

$m=-\frac{2}{3} $ gives min area value$$

$\therefore $ Min Area$ =6$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

Ekman said:
For the first part sub in y=mx+b into the ellipse equation and make discriminant = 0:

$4x^2 + 9y^2=36$

$x^2(4+9m^2) + 18mbx+9b^2-36 =0$

$324m^2b^2-4(4+9m^2)(9b^2-36)=0$

$\therefore b^2=9m^2+4$

$y=b $ and $ x=\frac{-b}{m}$

$\therefore Area=\frac{-b^2}{2m}$

$=\frac{-9m^2-4}{2m}$

$m=-\frac{2}{3} $ gives min area value$$

$\therefore $ Min Area$ =6$

howd you find area

Ekman · Feb 13, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION:

$$P is a point on the ellipse with equation $ \frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1 $ and focus S. The normal at P intersects the x-axis at Q. Show that $ QS=ePS$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

Ekman said:
NEXT QUESTION:

$$P is a point on the ellipse with equation $ \frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1 $ and focus S. The normal at P intersects the x-axis at Q. Show that $ QS=ePS$

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

$Apply implicit differinciation and inverse y' and multiply by -1 to get gradient of normal$

$y'=\frac{ya^2}{xb^2} = -1/y'$

$$ By finding the equation of the normal it is found that $ e^2x=x_0$

$\therefore $ distance formula with QS to get $ e^2-ae $ and $ ePS $ which is also $e^2-ae$
Hate coding on my phone

Ekman · Feb 13, 2015

Re: HSC 2015 4U Marathon

$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

dan964 · Feb 15, 2015

Re: HSC 2015 4U Marathon

dan964 said:
You do realise that $$ \alpha\beta\gamma\delta=1$ from product of roots
Anyway next question.

$$ Find the limiting sum of $ cis (x) + cis (2x) + cis (3x) + cis (4x) + ...$
$$ and hence simplify $ sin (x) + sin (2x) + sin (3x) + sin (4x)+ ...$

yep the correct answer is nope you can't find the limiting sum (what was I thinking?)

dan964 · Feb 15, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ [B]and the rhombus diagonals are along the coordinate axes[/B]. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

Since the diagonals of the rhombus are at 90 degrees
Possibly use the chord of contact is my guess that is of the form y=k or x=a.

Edit:
Find the vertices of the rhombus I would suggest.
To do this you need to find a tangent from (x1, y1) and by symmetry (-x1, y1)

Drsoccerball · Feb 15, 2015

Re: HSC 2015 4U Marathon

dan964 said:
Since the diagonals of the rhombus are at 90 degrees
Possibly use the chord of contact is my guess that is of the form y=0.
Find the vertices of the rhombus I would suggest.

Whats the chord of contact..

braintic · Feb 15, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Whats the chord of contact..

You haven't done 3U Parameters yet?

Drsoccerball · Feb 15, 2015

Re: HSC 2015 4U Marathon

braintic said:
You haven't done 3U Parameters yet?

my teacher said it wasnt important so we never learnt it properly...

Ekman · Feb 15, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

Don't be afraid to use parametric to solve it, however I suggest using Cartesian coordinates since it makes everything a lot more quicker.

braintic · Feb 15, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
my teacher said it wasnt important so we never learnt it properly...

Not important?? Tell your teacher they are an idiot.

Drsoccerball · Feb 15, 2015

Re: HSC 2015 4U Marathon

braintic said:
Not important?? Tell your teacher they are an idiot.

Nah he probably taught it.. its probably me not paying attention

Ekman · Feb 15, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Nah he probably taught it.. its probably me not paying attention

Why wouldn't you pay attention in class....

braintic · Feb 15, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Why wouldn't you pay attention in class....

I can think of good reasons. But if those reasons are correct, I sure hope he goes to a co-ed school.

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