Author Topic: I need help with question 15 &16 please!! (Read 1201 times) Tweet Share

RuiAce · « **Reply #1 on:** January 31, 2017, 08:34:29 pm »

Important warning: The Extension questions in Cambridge are designed purely for the sake of self-interest. They are only for you to build superior techniques when you feel necessary. They do not reflect the scope of the 2U, 3U or 4U syllabuses and therefore I will not, in general, do these questions from here on after.
$\text{Note that the line }\boxed{y-a^2=m(x-a)}\\ \text{is just direct substitution into the point-gradient formula }y-y_1=m(x-x_1)\\ \text{Effectively, we want to solve for points of intersection with }\boxed{y=x^2}$
$\text{By substituting:}\\ \begin{align*}x^2-a^2&=m(x-a)\\ (x-a)(x+a)&=m(x-a)\\ (x-a)(x+a)-m(x-a)&=0\\ (x-a)[(x+a)-m]&=0\\ \therefore x&=a, \, m-a\end{align*}$
$\text{if the points should coincide, then we simply equate the two:}\\ a=m-a \implies m=2a$
$\text{Now, we must turn back to the geometrical significance}\\ \text{of what it means for the points to }\textbf{coincide}.$
GeoGebra simulation attached
$\text{The parabola }y=x^2\text{ and the line }y-a^2=m(x-a)\\ \text{have been plotted together}\\ \text{Because }a\text{ is unknown, it is treated as a slider.}\\ \text{Experiment with multiple values of }a\text{ via the slider.}$
$\text{The point of the GeoGebra simulation is}\\ \text{No matter what value of }a\text{ you pick}\\ \text{If }m\neq 2a\text{, then the line cuts the parabola twice.}$
$\text{But instead, if you set }m\text{ to EQUAL to }2a\\ \text{You only get one point of intersection.}\\ \text{In fact. When }m=2a\text{, the line is }\textbf{tangent}\text{ to the parabola}$
$\text{This is because, in the original question, }m=2a\text{ is when we cause the}\\ \textbf{two points of intersection to coincide.}\\ \text{Because they are COINCIDING, we ultimately have only one point-of-intersection.}$
The GeoGebra simulation is given with a=1 and m=1. Leaving a=1, we would require m=2 to see this happen.
$\text{So ultimately, if }m=2a\text{ then the line}\\ \text{becomes tangent to the parabola.}\\ \text{But what is }m?$
$\boxed{m\text{ is nothing more than the }\textbf{gradient of the line!}}\\ \text{Hence, if our point is at }P(a,a^2)\\ \text{i.e. }\boxed{x=a}\\ \text{and we know that the GRADIENT of the tangent is }\boxed{m=2a}\\ \text{We ultimately deduce that }m=\frac{dy}{dx}=2a=2x$
_________________________________________
$\text{Q16 follows a similar process, except}\\ \text{the two points of intersection to equate will be}\\ x=\frac{m\pm \sqrt{m^2+4b}}2$

Keerit · « **Reply #2 on:** January 31, 2017, 08:35:32 pm »

Thank you !

QC · « **Reply #3 on:** February 05, 2017, 02:17:40 pm »

Quote from: RuiAce on January 31, 2017, 08:34:29 pm

Important warning: The Extension questions in Cambridge are designed purely for the sake of self-interest. They are only for you to build superior techniques when you feel necessary. They do not reflect the scope of the 2U, 3U or 4U syllabuses and therefore I will not, in general, do these questions from here on after.
$\text{Note that the line }\boxed{y-a^2=m(x-a)}\\ \text{is just direct substitution into the point-gradient formula }y-y_1=m(x-x_1)\\ \text{Effectively, we want to solve for points of intersection with }\boxed{y=x^2}$
$\text{By substituting:}\\ \begin{align*}x^2-a^2&=m(x-a)\\ (x-a)(x+a)&=m(x-a)\\ (x-a)(x+a)-m(x-a)&=0\\ (x-a)[(x+a)-m]&=0\\ \therefore x&=a, \, m-a\end{align*}$
$\text{if the points should coincide, then we simply equate the two:}\\ a=m-a \implies m=2a$
$\text{Now, we must turn back to the geometrical significance}\\ \text{of what it means for the points to }\textbf{coincide}.$
GeoGebra simulation attached
$\text{The parabola }y=x^2\text{ and the line }y-a^2=m(x-a)\\ \text{have been plotted together}\\ \text{Because }a\text{ is unknown, it is treated as a slider.}\\ \text{Experiment with multiple values of }a\text{ via the slider.}$
$\text{The point of the GeoGebra simulation is}\\ \text{No matter what value of }a\text{ you pick}\\ \text{If }m\neq 2a\text{, then the line cuts the parabola twice.}$
$\text{But instead, if you set }m\text{ to EQUAL to }2a\\ \text{You only get one point of intersection.}\\ \text{In fact. When }m=2a\text{, the line is }\textbf{tangent}\text{ to the parabola}$
$\text{This is because, in the original question, }m=2a\text{ is when we cause the}\\ \textbf{two points of intersection to coincide.}\\ \text{Because they are COINCIDING, we ultimately have only one point-of-intersection.}$
The GeoGebra simulation is given with a=1 and m=1. Leaving a=1, we would require m=2 to see this happen.
$\text{So ultimately, if }m=2a\text{ then the line}\\ \text{becomes tangent to the parabola.}\\ \text{But what is }m?$
$\boxed{m\text{ is nothing more than the }\textbf{gradient of the line!}}\\ \text{Hence, if our point is at }P(a,a^2)\\ \text{i.e. }\boxed{x=a}\\ \text{and we know that the GRADIENT of the tangent is }\boxed{m=2a}\\ \text{We ultimately deduce that }m=\frac{dy}{dx}=2a=2x$
_________________________________________
$\text{Q16 follows a similar process, except}\\ \text{the two points of intersection to equate will be}\\ x=\frac{m\pm \sqrt{m^2+4b}}2$

But next minute an ext question from Cambridge 3u is the last question of the 4u paper. The derangement question was one of the ext questions in the perms and combs section of yr 12 Cambridge 3u

RuiAce · « **Reply #4 on:** February 05, 2017, 10:18:11 pm »

Quote from: aroon217 on February 05, 2017, 02:17:40 pm

But next minute an ext question from Cambridge 3u is the last question of the 4u paper. The derangement question was one of the ext questions in the perms and combs section of yr 12 Cambridge 3u

Well I'll share a quick rant. I am most certainly fine with them introducing new terminology in an exam paper - they clearly explain what a derangement is. But why the hell was a recurrence relation question sitting there? In what part of the 4U course were students taught how to handle recurrence in perms and combs? The induction question was fine (albeit crazy in that it was only worth 2 marks), and I'm not surprised by recurrence in series (definition of an A.P.), but here? I had nothing to say - at the time I wrote the solutions, all I could think of was why is there something I learnt a mere 1.5 months ago floating around here

Sure, the HSC has been getting dumber. But spiking the difficulty like this is ridiculous in my opinion. I get it, in 2015 they messed up and put a no-brainer part ii) in the perms and combs question. But really? You choose to go from something easy to something insane? Feels like a repeat of the year 2013 with 2U and 3U.

I don't have a problem with them using a textbook question in the exam. Exam writers run out of material eventually, and if you did it before then you hit the jacpot. I'm more fussed about putting something which I don't feel properly belongs in the scope of the course there.

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Author Topic: I need help with question 15 &16 please!! (Read 1201 times) Tweet Share

Keerit

I need help with question 15 &16 please!!

RuiAce

Re: I need help with question 15 &16 please!!

Keerit

Re: I need help with question 15 &16 please!!

QC

Re: I need help with question 15 &16 please!!

RuiAce

Re: I need help with question 15 &16 please!!

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