2008 HSC question 8(b) (1 Viewer)

[lukezy]

Another question for you guys since you were so helpful with the last one =]

Usually question 8(b) is a rite-off but this one was actually quite straight forward.
I'm fine with parts i-iii, but iv threw me off a bit.

Heres the link for the paper
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext2_05.pdf

The worked solution for iv reads....
CP x PD = p(q+r)
similarly
QD x QC = q(p+r)
CP x PD = QD x QC (since C, P, Q and D are collinear and alpha remainds constant) *
So p(q+r) = q(p+r)
pq + pr = pq+qr or pr = qr
therefore p = q

when stating that CP x PD = QD x QC are they not assuming what we are trying to prove?

Say hypothetically that DQ = 2, QP = 5 and PC = 3
They are still all collinear and alpha is still constant
but when subbed into this "rule" we get that 21 = 16
I cannot see the logic in this solution

Any help would be appreciated
Thankyou ^^

 

[tacogym27101990]

you have the 2008 hsc paper??
care to share??

 

[conics2008]

man 2005 hsc paper was soo hard, I really feel sorry for the people who did the 2005 paper, it was hard ass !!

any one find the difference between 2005 and 2007..

2007 was alot easier =]

 

[samwell]

Hahahahah i thought it was the '08 paper i was wow this guy has had like a vision on the exam. lol u mean 2005 HSC lol.

 

[lukezy]

okok 2005 hsc paper

guess the stress of the hsc has really got to me hahaha

anyway, no ideas anyone?

 

[cheney31]

you have the 2008 hsc paper??
care to share??

LOL....

 

[lukezy]

bump*.....
so no1 knows how to explain it? :S

 

[lolokay]

no idea what their reasoning is, but what I would say (is this appropriate reasoning?) is that:
in a hyperbola such as x²/a² - y²/a² = 1, then if the distances of the intervals of a line intersecting the hyperbola and the corresponding assymptote at 2 points, is equal, then the y components of each interval will be the same (since the intervals have the same gradient). So, if the y axis of the graph is scaled so that we have the graph x²/a² - y²/b² = 1, then intervals (p and q) will still be the same.

Now suppose we were to scale and rotate the hyperbola x²/a² - y²/a² = 1 so that it became the hyperbola xy = 1 and intersect it with the line y = mx+b. Then, if the lengths of the intervals of the line where it intersects the hyperbola and its asymptotes (the x and y axis) are equal, then it must be the case that the intervals on x²/a² - y²/b² = 1 (p and q) are also equal

the line y = mx + b intersects the hyperbola xy = 1, y = 1/x at mx+b = 1/x
solving the quadratic gives x = (-b+-sqrt[b² + 4m])/2m
the line y = mx + b intersects the x and y axis at x=0 and x=-b/m
so the lengths of the intervals in the x direction are:
(-b+sqrt[b² + 4m])/2m
and -2b/2m - -b/2m - -sqrt[b² + 4m])/2m
= (-b+sqrt[b² + 4m])/2m

the intervals are equal, and thus p=q


goddamn that's long and confusing to write out.. can someone comment on the quality of this proof?

 

[shaon0]

no idea what their reasoning is, but what I would say (is this appropriate reasoning?) is that:
in a hyperbola such as x²/a² - y²/a² = 1, then if the distances of the intervals of a line intersecting the hyperbola and the corresponding assymptote at 2 points, is equal, then the y components of each interval will be the same (since the intervals have the same gradient). So, if the graph if the y axis of the graph is scaled so that we have the graph x²/a² - y²/b² = 1, then intervals (p and q) will still be the same.

Now suppose we were to scale and rotate the hyperbola x²/a² - y²/a² = 1 so that it became the hyperbola xy = 1 and intersect it with the line y = mx+b. Then, if the lengths of the intervals of the line where it intersects the hyperbola and its assymptotes (the x and y axis) are equal, then it must be the case that the intervals on x²/a² - y²/b² = 1 (p and q) are also equal

the line y = mx + b intersects the hyperbola xy = 1, y = 1/x at mx+b = 1/x
solving the quadratic gives x = (-b+-sqrt[b² + 4m])/2m
the line y = mx + b intersects the x and y axis at x=0 and x=-b/m
so the lengths of the intervals in the x direction are:
(-b+sqrt[b² + 4m])/2m
and -2b/2m - -b/2m - -sqrt[b² + 4m])/2m
= (-b+sqrt[b² + 4m])/2m

the intervals are equal, and thus p=q


goddamn that's long and confusing to write out.. can someone comment on the quality of this proof?

lol....when are they taking conics out (ie. geometry)

 

[lukezy]

ummm.... ill read over it a few times and see if i can decipher it

thanks heaps btw

Worst comes to worst ill just show my tutor your solution and see what he thinks of it. But atleast i can see logic in it rather than the shitty solution in my past hsc papers book.

Still open for new explanations