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I got c , but the answers says b. Not sure why?
Any one have any clue??
http://imgur.com/a/3ZNPJ
12a & 13a? (Chapter 9C)
For 12a
The equation of the parabola is in the form
The tangent has the equation
Sub that into the parabola to get a quadratic in terms of x and a, then use the discriminate and let that equal to 0 (since it's a tangent, so there's only one solution) to solve for a. Then sub that value back into the first equation.
“Smart people learn from their mistakes. But the real sharp ones learn from the mistakes of others.”
― Brandon Mull
Make sure the tangent is effectively portrayed
Not sure if directrix on the diagonal is covered in Maths Extension 1. I did draw a graph of the points and am aware that the formula from the Focal Point to the straight line is:
Im not sure if the parabola is parallel to the x-axis or parallel to the y-axis.
Here was the diagram I tried to work from :
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13a
there might be a shorter way
Last edited by jathu123; 13 Sep 2016 at 11:41 PM.
2017
4u99 - 3u98 - EngAdv88 - Phys94 - Chem94
Atar : 99.55
Course: Engineering(Hons)/Commerce @ UNSW
ATAR is just a number
Think its the first time I have done a question that involved a parabola with a diagonal directrix. (Don't think I have seen too many of those type in past exam papers)
Think I also have to memorise the distance of a point on a parabola to the focus is equivalent to point on a parabola to the directrix. (PS=PM)
When I looked at your final equation, without using a calculator my initial thoughts were it a circle or hyperbola.
But drawing it on desmos it looked like that (not sure how would you describe that graph).
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Q 7 (b) EX 9A
I just wondering has there been a mistake in this answer of the 3U Year 11 textbook?
ii) ==> PA² + PB² + PC² = 3x² + 3y² - 14x - 6y + 78 = 77
==> 3x² + 3y² - 14x - 6y = -1
==> x² + y² - 14x/3 - 2y = -1/3
Grouping, (x² - 14x/3) + (y² - 2y) = -1/3
==> (x² - 14x/3 + 49/9) + (y² - 2y + 1) = -1/3 + 49/9 + 1
==> (x - 7/3)² + (y - 1)² = 55/9
The back of the book has the answer as (2,1) for centre and radius of
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ATAR: 99.75
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2017
4u99 - 3u98 - EngAdv88 - Phys94 - Chem94
Atar : 99.55
Course: Engineering(Hons)/Commerce @ UNSW
ATAR is just a number
OK I decided to redo Q 7 a and 7b from EX 9A
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Buy my books/notes cheaply here!
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Uni Course: Actuarial Studies and Statistics at MQ -- PM me if you have questions
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ATAR: 99.75
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Hi, need help with Exercise 8H Page 311 Extension Question 27.
If alpha and beta are the roots of x^2 = 5x - 8. Find (a)^1/3 + (b)^1/3 without finding the roots.
Tried cubing (a^1/3 + b^1/3) but got stuck.
Any help would be appreciated.
Also, with Extension question 26 (same page),
If the equations mx^2 + 2x +1 =0 and x^2 + 2x +m = 0 have a common root, find the possible values of m and the value of the common root in each case. I let first equation be p(x) and the second be f(x) and the common root be alpha.
If alpha is the common root p(a) = f(a) = 0 and hence you get,
a^2(m-1) + (1-m) = 0, and that the common root a is x=-1 when m = 1
I had no idea how to get the second common root of the two equations (its x = 1 when m = -3),
So I used product of roots of the above equation a^2(m-1) + (1-m) = 0 as a quadratic is a, and the product of roots is thus -1
Hence I found the other root to be x = -1 and subbing x = -1 into any of the given functions p(x), f(x) and letting it equal 0 I then got m = -3.
Is there a way to find the 2nd m value prior to finding the 2nd common root?
(because this is what the question seems to imply find m then find the common root)
Sorry for long post couldn't find anything on either of the questions
Thanks!
Last edited by Drongoski; 11 Oct 2016 at 10:48 PM.
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Yes, I got that part lol.
But what would you do to get a^(1/3) + b^(1/3) from a+b = 5 and ab = 8
What I tried was,
(a^(1/3) + b^(1/3))^3 and got a + 3a^(2/3)b^(1/3) + 3a^(1/3)b^(1/3) + b
Tried doing something with the 3a^(2/3)b^(1/3) + 3a^(1/3)b^(2/3) but ended up getting more a^(1/3) + b^(1/3) terms when I tried factorising by taking out powers of ab
Or would cubing a^(1/3) + b^(1/3) be useless for this question?
Thank you so much
Last edited by Drongoski; 12 Oct 2016 at 1:18 PM.
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EX7D Q 9 (b) is there a mistake with the answer?
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One of the angles I believe to be right (the 71 34') and the other I believe to be wrong. Would love it , if somebody could confirm.
Seeing that the two points of intersection are at x=2, -3
I then substituted this back into the derivative which is dy/dx=2x-1=tan
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The 2 angles in degrees: 71.565 and 98.13
being: inverse tan (3) and 180 - inverse tan (7)
Maybe my calculator is exceptionally accurate.
Last edited by Drongoski; 25 Oct 2016 at 11:00 PM.
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I had an attempt at the question and I can't quite get it
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My reasoning as to why I subtract 180 from 81'52' is that the tan theta value is a value of negative 7, which lies somewhere in the 2nd quadrant of the ASTC diagram.
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