HSC 2015 MX2 Marathon (archive) (3 Viewers)

SammyT123 · Feb 24, 2015

Re: HSC 2015 4U Marathon

How do I sketch y=logx/x

When I get to the y' step, I let
U = logx
Du=1/x dx

Y'=udu

This is somehow wrong :/

Sent from my iPhone using Tapatalk

Drsoccerball · Feb 24, 2015

Re: HSC 2015 4U Marathon

SammyT123 said:
How do I sketch y=logx/x

When I get to the y' step, I let
U = logx
Du=1/x dx

Y'=udu

This is somehow wrong :/

Sent from my iPhone using Tapatalk

sketch logx and x and divide every logx ordinate by x ordinate

Chlee1998 · Feb 24, 2015

Re: HSC 2015 4U Marathon

In triangle ABC, D is a point on BC. the incircles of triangle ADB and ADC are both tangent to AD at the same point. Show that the incircle of ABC touches BC at D

FrankXie · Feb 25, 2015

Re: HSC 2015 4U Marathon

Chlee1998 said:
In triangle ABC, D is a point on BC. the incircles of triangle ADB and ADC are both tangent to AD at the same point. Show that the incircle of ABC touches BC at D

Let the incircles of triangle ADB and ADC both touch AD at the point E. (given)
Then DE=(AD+BD-AB)/2, DE=(AD+CD-AC)/2. (use "two tangents through an external point are equal" several times)
Therefore, AB-BD=AC-CD. (manipulate simple algebra on the above equations)
This forces that the incircle of ABC touch BC at D. (if otherwise, they touch at some other point X, then AB-BX=AC-CX, and we have BD-CD=BX-CX, noting that both D and X are on the line BC, it follows that points D and X must be identical.)

FrankXie · Feb 25, 2015

Re: HSC 2015 4U Marathon

new question

$The equations of two ellipses are (i) $ 4x^2+9y^2=36 $, (ii) $ 2x^2+3y^2=30 $. The tangent to ellipse (i) at the point $ R(3\cos\theta,2\sin\theta) $ meets the ellipse (ii) at the points $ P $ and $ Q $. Show that the tangents at $ P $ and $ Q $ to ellipse (ii) are at right angles to one another.$

Drsoccerball · Feb 25, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$The equations of two ellipses are (i) $ 4x^2+9y^2=36 $, (ii) $ 2x^2+3y^2=30 $. The tangent to ellipse (i) at the point $ R(3\cos\theta,2\sin\theta) $ meets the ellipse (ii) at the points $ P $ and $ Q $. Show that the tangents at $ P $ and $ Q $ to ellipse (ii) are at right angles to one another.$

I tried it and got a very big quadratic but after you get the quadratic sub it into ii) y' and multiply to show =-1 at school atm

FrankXie · Feb 25, 2015

Re: HSC 2015 4U Marathon

next one

$Find the locus of all points through which two tangents drawn to the ellipse $ \frac{x^2}{16}+\frac{y^2}{9}=1 $ are perpendicular to each other.$

VBN2470 · Feb 25, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
next one

$Find the locus of all points through which two tangents drawn to the ellipse $ \frac{x^2}{16}+\frac{y^2}{9}=1 $ are perpendicular to each other.$

$Using properties of tangents, discriminants and product of roots, we get $x^2+y^2=25$ i.e. the locus describes a circle centred at $(0,0)$, with radius 5 units.$

Soulful · Mar 1, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$The equations of two ellipses are (i) $ 4x^2+9y^2=36 $, (ii) $ 2x^2+3y^2=30 $. The tangent to ellipse (i) at the point $ R(3\cos\theta,2\sin\theta) $ meets the ellipse (ii) at the points $ P $ and $ Q $. Show that the tangents at $ P $ and $ Q $ to ellipse (ii) are at right angles to one another.$

That was a fun question!

At first I tried to brute force it with algebra but that failed miserably. Then it struck me that using a polynomials approach would get the proof much more elegantly and much faster as well

Will post my answer as soon as I remember how to use latex again haha

Soulful · Mar 1, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$The equations of two ellipses are (i) $ 4x^2+9y^2=36 $, (ii) $ 2x^2+3y^2=30 $. The tangent to ellipse (i) at the point $ R(3\cos\theta,2\sin\theta) $ meets the ellipse (ii) at the points $ P $ and $ Q $. Show that the tangents at $ P $ and $ Q $ to ellipse (ii) are at right angles to one another.$

$Let points P and Q be $ (x_{1},y_{1}) $ and $ (x_{2}, y_{2}) $ respectively.$

$From implicitly differentiating $, m_{P} $ and $ m_{Q} $ are $ \frac{-2x_{1}}{3y_{1}} $ and $ \frac{-2x_{2}}{3y_{2}} $ respectively.$

$$ If the tangents from P and Q meet at right angles, then $ m_{Q}m_{P} = -1$

$\therefore \frac{4x_{1}x_{2}}{9y_{1}y_{2}} = -1 $(R.T.P) $$

$$ The tangent to (i) at R has equation $ 2x\cos\theta + 3y\sin\theta = 6$

$x = \frac{6-3y\sin\theta}{2\cos\theta} $ Substitute this into (ii) $$

$36-36y\sin\theta + 9y^2\sin^2\theta + 6y^2\cos^2\theta - 60\cos^2\theta = 0$

$$ The roots of this equation must be $ y_{1} $ and $ y_{2}$

$\therefore y_{1}y_{2} = \frac{36 - 60\cos^2\theta}{9\sin^2\theta + 6\cos^2\theta} = \frac{20\sin^2\theta - 8}{2 + \sin^2\theta}$

$Similarly, for x$

$x_{1}x_{2} = \frac{36-90\sin^2\theta}{6\sin^2\theta+\cos^2\theta} = \frac{18-45\sin^2\theta}{2+\sin^2\theta}$

$\therefore \frac{4x_{1}x_{2}}{9y_{1}y_{2}} = 4(\frac{18-45\sin^2\theta}{2+\sin^2\theta}) \div 9(\frac{20\sin^2\theta - 8}{2 + \sin^2\theta}) = \frac{72 - 180\sin^2\theta}{180\sin^2\theta-72} = -1$

$$ Consequently the tangents are at right angles to each other$

FML that took way too long to turn into latex

Ekman · Mar 1, 2015

Re: HSC 2015 4U Marathon

Interesting approach. Here's my solution:

$Equation of tangent = $ \frac{x\cos\theta}{3} + \frac{y\sin\theta}{2} $ is also the equation of the chord of contact of ii)$

$\therefore $ let the external point of the chord of contact of ii) be $ (x_1,y_1)$

$\therefore \frac{xx_1}{15} + \frac{yy_1}{10}=1 $ is the chord of contact of ii) $$

$\therefore $ Through equating coefficients $ x_1=5\cos\theta $ and $ y_1=5\sin\theta $ ( Thus the equation of the locus is a circle : $ x^2+y^2=25 $ )$$

$$ Let the equation of the tangent to ii) be $ y=mx+c$

$\therefore $ The discriminant of the equation of tangent subbed into the equation of ii) is: $ c^2=15m^2+10$

$5\sin\theta - m 5\cos\theta = c $ (As the external point fits into the equation of tangent) $$

$(5\sin\theta - m 5\cos\theta)^2 = 15m^2+10$

$\therefore $ Product of roots = $ \frac{25\sin^2\theta -10}{25\cos^2\theta - 15} = \frac{25\sin^2\theta -10}{10-25\sin^2\theta} = -1$

$\therefore$ They are perpendicular $$

Ekman · Mar 2, 2015

Re: HSC 2015 4U Marathon

$$A normal is drawn from P $(cp,\frac{c}{p}) $ from the hyperbola $ xy=c^2 $ and it cuts the hyperbola again at Q $(cq,\frac{c}{q}).$
$Let M be the midpoint of PQ.$
$$Therefore show the locus of M is: $ c^2(x^2-y^2)^2+4x^3y^3=0$

braintic · Mar 2, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$A normal is drawn from P $(cp,\frac{c}{p}) $ from the hyperbola $ xy=c^2 $ and it cuts the hyperbola again at Q $(cq,\frac{c}{q}).$
$Let M be the midpoint of PQ.$
$$Therefore show the locus of M is: $ c^2(x^2-y^2)^2+4x^3y^3=0$

And sketch this locus!

Ekman · Mar 2, 2015

Re: HSC 2015 4U Marathon

braintic said:
And sketch this locus!

Nice...

Kaido · Mar 3, 2015

Re: HSC 2015 4U Marathon

yeah so many people get the locus part wrong, esp. the restricted domain in which the locus exists. Def draw a diag...

Axio · Mar 7, 2015

Re: HSC 2015 4U Marathon

$$Using the substitution:$ \ u= sec\alpha + tan\alpha, \ $solve the integral$ \ \int sec\alpha d\alpha$

FrankXie · Mar 7, 2015

Re: HSC 2015 4U Marathon

Axio said:
$$Using the substitution:$ \ u= sec\alpha + tan\alpha, \ $solve the integral$ \ \int sec\alpha d\alpha$

$\sec\alpha=\frac{\sec^2\alpha+\sec\alpha\tan\alpha}{\sec\alpha+\tan\alpha}$

Drsoccerball · Mar 7, 2015

Re: HSC 2015 4U Marathon

Trying to make a question again so this may turn out bad but:
There are 5 roads to get to a destination. 3 are connected and the other two are not connected to any of the roads. If the probability of a road not being constructed yet is p.
I) whats the probability of the not making it to your destination
ii) if there is 15 suburbs with identical streets as i) what is the probability of not making it to your destination.

Drsoccerball · Mar 7, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Trying to make a question again so this may turn out bad but:
There are 5 roads to get to a destination. 3 are connected and the other two are not connected to any of the roads. If the probability of a road not being constructed yet is p.
I) whats the probability of the not making it to your destination
ii) if there is 15 suburbs with identical streets as i) what is the probability of not making it to your destination.

EDIT: ignore ii... I need to provide a diagram

braintic · Mar 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Trying to make a question again so this may turn out bad but:
There are 5 roads to get to a destination. 3 are connected and the other two are not connected to any of the roads. If the probability of a road not being constructed yet is p.
I) whats the probability of the not making it to your destination
ii) if there is 15 suburbs with identical streets as i) what is the probability of not making it to your destination.

Connected to what?

HSC 2015 MX2 Marathon (archive) (3 Viewers)

Active Member

Well-Known Member

Member

Active Member

Active Member

Well-Known Member

Active Member

Well-Known Member

HSC Hipster

HSC Hipster

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

be.

=o

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)