HSC 2016 MX1 Marathon (archive) (3 Viewers)

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

leehuan said:
The parabola takes the form (x-2)^2=4a(y+1)

A wolfram alpha approach: Sub y=2x-7 in and simplify the expression

Take the quadratic discriminant, set it equal to 0 and take your value for a

So you say let y=2x-7 (which I didn't get at the time)

makes it :

$(x-2)^2=4a [(2x-7)+1]$

$x^2-4x+4=4a(2x-6)$

$x^2-x(4-8a)+(4-24a)=0$

$\triangle =b^2-4ac=0$

$(4-8a)^2-4(4-24a)=0$

$16-64a+64a^2-16+96a=0$

$64a^2+32a=0 \implies 32a(2a+1)=0$

$a=\frac{-1}{2}$

a cannot be 0 in this case because it must be a quadratic form.

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Find the equation of the parabola whose axis is parallel to the y-axis, vertex is (2,-1) and has a tangent with equation y=2x-7.

I have no answer to compare against . This question (can't remember off the top of my head) but was used in the Trials at one of the schools in NSW.

Taking the standard form

4a(y+1) = (x-2)²

The parametric form of the point P(x,y) is P(2at+2, at²-1), with unaltered gradient t

The tangent is thus y + 1 = tx - 2t - at²

Equating tangents yields: at² + 2t + 1 = 7, t = 2

4a + 4 + 1 = 7

4a = 2

a = 1⁄2

Hence, 2(y+1) = (x-2)²

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

I must admit I did it a slightly different way in my attempt

$y=k(x-2)^2-1$

$$ I then did the derivative $ \frac{dy}{dx}=2k(x-2)$

$$ At x=0,$ \frac{dy}{dx}=2$

$2=2k(0-2)$

$\therefore k=\frac{-1}{2}$

$y=\frac{-1}{2} (x-2)^2-1 \implies 2y=-(x-2)^2-2$

$2y+2=-(x-2)^2$

$2(y+1)=-(x-2)^2$

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
I must admit I did it a slightly different way in my attempt

$y=k(x-2)^2-1$

$$ I then did the derivative $ \frac{dy}{dx}=2k(x-2)$

$$ At x=0,$ \frac{dy}{dx}=2$

$2=2k(0-2)$

$\therefore k=\frac{-1}{2}$

$y=\frac{-1}{2} (x-2)^2-1 \implies 2y=-(x-2)^2-2$

$2y+2=-(x-2)^2$

$2(y+1)=-(x-2)^2$

You're doing something wrong because Geogebra confirms my answer is the correct one.

leehuan · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
So you say let y=2x-7 (which I didn't get at the time)

makes it :

$(x-2)^2=4a [(2x-7)+1]$

$x^2-4x+4=4a(2x-6)$

$x^2-x(4-8a)+(4-24a)=0$

$\triangle =b^2-4ac=0$

$(4-8a)^2-4(4-24a)=0$

$16-64a+64a^2-16+96a=0$

$64a^2+32a=0 \implies 32a(2a+1)=0$

$a=\frac{-1}{2}$

a cannot be 0 in this case because it must be a quadratic form.

Somewhat like that. (I didn't do the question so I can't say if the final answer is correct.)

Also let y=2x-7 was just solving simultaneous equations. Don't forget a tangent to a conic only meets it once therefore 1 unique pt of intersection.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Taking the standard form

4a(y+1) = (x-2)²

The parametric form of the point P(x,y) is P(2at+2, at²-1), with unaltered gradient t

The tangent is thus y + 1 = tx - 2t - at²

Is this based on the formula

$y-y_1=m(x-x_1)$ ?

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Is this based on the formula

$y-y_1=m(x-x_1)$ ?

Yes, however, I have rearranged it slightly.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Taking the standard form

4a(y+1) = (x-2)²

The parametric form of the point P(x,y) is P(2at+2, at²-1), with unaltered gradient t

The tangent is thus y + 1 = tx - 2t - at²

Equating tangents yields: at² + 2t + 1 = 7, t = 2

If Im right this is something we need to memorise right? P(2at+2, at²-1)

Also how did you substitute t=2?

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
If Im right this is something we need to memorise right? P(2at+2, at²-1)

Also how did you substitute t=2?

No, I am finding the parameter such that the tangents are equivalent. This can be seen by comparing the coefficients of x on both tangents.

Then the parameter value is distributed through the entire tangent to remove the variable, allowing the value of a to be found.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
No, I am finding the parameter such that the tangents are equivalent. This can be seen by comparing the coefficients of x on both tangents.

Then the parameter value is distributed through the entire tangent to remove the variable, allowing the value of a to be found.

OK I got it now, you were comparing coefficients.

I explained this question wrong to my student tonight.

Will email her with the right solution tomorrow.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

I do memorise for a parametric equation going through the vertex (0,0)

the parametric equations were

$x=2at \ y=at^2$

I will have to re-read my formulae book again in regards to shifts in vertices and how that affects the parametric equations.

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
OK I got it now, you were comparing coefficients.

I explained this question wrong to my student tonight.

Will email her with the right solution tomorrow.

I don't think there's a right solution. Both Parametric and Cartesian approaches are perfectly valid in my opinion.

davidgoes4wce · Jul 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
I don't think there's a right solution. Both Parametric and Cartesian approaches are perfectly valid in my opinion.

I just notice that a constant for yours way positive 1/2 , mine was -1/2 .

Paradoxica · Jul 12, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
I just notice that a constant for yours way positive 1/2 , mine was -1/2 .

Geogebra verifies the veracity of my equation.

InteGrand · Jul 15, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Wait what does RTP stand for?

How do you get from the 4th line to the 5th line in the proving by RHS? And how does the end result relate to the inequality the question is asking to prove? Sorry, I still don't get it.

RTP means "required to prove".

$\noindent To get from line 4 to line 5, leehuan used the result $ab \leq \frac{a^2+b^2}{2}$. This is true because $\left(a-b\right)^2 \geq 0$, and then expanding and rearranging this gives that result.$

InteGrand · Jul 15, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
And how does the end result relate to the inequality the question is asking to prove? Sorry, I still don't get it.

leehuan added a post afterwards, after some discussion on that Q. by others.

leehuan said:
Would not have thought about that at all. Just by substituting this result into line 3 of the proof the result falls out immediately.

$\begin{align*}(a-b)(a^k-b^k)&\ge 0\\ a^{k+1}+b^{k+1}& \ge ab^k+a^kb\end{align*}$

I was trying to think about why a^kb+ab^k≤a^k+1+b^k+1. Guess I've never seen that before to use it.

davidgoes4wce · Jul 15, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Taking the standard form

The parametric form of the point P(x,y) is P(2at+2, at²-1), with unaltered gradient t

This was from my formula sheet:

Now from what I gather is if its been shifted to the right, the parametric equation for x=2at+2.

If the equation has a vertical shift of 1 unit down, the parametric equation is y=at^2-1.

It's something that has never really crossed my mind when thinking of this in terms of parabolas. (or parametric equations in relation to parabolas)

Paradoxica · Jul 15, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
This was from my formula sheet:

Now from what I gather is if its been shifted to the right, the parametric equation for x=2at+2.

If the equation has a vertical shift of 1 unit down, the parametric equation is y=at^2-1.

It's something that has never really crossed my mind when thinking of this in terms of parabolas. (or parametric equations in relation to parabolas)

For simple parameters, the relation is easy to define in terms of co-ordinate transforms.

But for other things such as angular transforms... that is not as nice.

davidgoes4wce · Jul 17, 2016

Re: HSC 2016 3U Marathon

Q9 from Cambridge text and answer.

Is it ok for Q 9 to say alternate angles instead of corresponding angles as an explanation for the rhombus ?

I wanted to say something along the lines of:

$\angle ABE=\angle CDE $ (alternate angles) $$

$\angle BAE=\angle DCE $ (alternate angles) $$

$\therefore AB \parallel CD.$

$\angle ADE=\angle CBE $ (alternate angles) $$

$\angle DAE=\angle BCE $ (alternate angles) $$

$\therefore AD \parallel CB$

My only thinking of using the word (corresponding) in this example is they are 4 separate triangles within the quadrilateral. But its something im unsure of.

InteGrand · Jul 17, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Q9 from Cambridge text and answer.

Is it ok for Q 9 to say alternate angles instead of corresponding angles as an explanation for the rhombus ?

I wanted to say something along the lines of:

$\angle ABE=\angle CDE $ (alternate angles) $$

$\angle BAE=\angle DCE $ (alternate angles) $$

$\therefore AB \parallel CD.$

$\angle ADE=\angle CBE $ (alternate angles) $$

$\angle DAE=\angle BCE $ (alternate angles) $$

$\therefore AD \parallel CB$

My only thinking of using the word (corresponding) in this example is they are 4 separate triangles within the quadrilateral. But its something im unsure of.

If we said alternate angles as the reason for the equality, that'd mean we already know the lines are parallel, but this is what we're trying to show (we don't know it yet). We use corresponding angles in congruent triangles (proof of congruence done in first part of the question) to justify the equality, because we already know about the congruence.

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