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oh i didnt see that - thanks

omg i cant believe i knew that this thread was started by you b4 opening it. It just has that weird essence to it.

stuff completing the square unless the question asks, if the question did ask just swear at your teacher - also if you complete the square how do you get a circle.

this question is stuffing you up ehhh ;)

Thats not a bloody circle cos coeffs of x and y are different PA^2 = (X+6)^2 + (Y-5)^2 = X^2 + 12X + Y^2 -10Y + 61 PB^2 = (X-3)^2 + (Y+1)^2 = X^2 - 6X + Y^2 + 2Y + 10 4PA^2 = 9B^2 Then i get 5x^2 -102x + 5y^2 +68y = 154 WHICH IS A CIRCLE SINCE COEFFS OF X AND Y ARE SAME question...

hey realise thanks for the suggestion. ive already completed that question lol - its in our 4unit booklet. but the question specifies that A (or P') is directly above P and Q' is directly below Q. Or should I just consider the triangles given in the question and use that to show the distances...

thanks :) ... but wouldnt it take a few weeks for you to learn hyperbola properties (assuming you learn ellipse first)

Prove that if chord PQ cuts hyperbola x^2/a^2 - y^2/b^2 = 1 at P and Q and the asymptotes at P' and Q' then PP' = QQ' Can someone please show me how to do this. I tried both parametric and cartesian approach. And using distance formula gets way too complicated as there are like 3 or 4 terms...

yh it is but my class learnt it in three unit for fun lol

and q12 - max area of rectangle occurs when all vertices on circumfgerence - using circle geo it can be shown the the centre of the rectangle (where diagonals intersect ) is the centre of the circle. Therefore diagonal = diameter = 12cm let dimensions of rect be xcm by ycm using pythagoras x^2...

for the surfboard question let the length of the rectangle part be X and the height be 2R ---> thus the radius of the semicircle is R Perimeter = X + X + 2R + 1/2 (2piR) = 2X + R(pi+2) = 4 From this X = 1/2 [ 4 - R(pi + 2)] Area of the shape A = 2RX + 1/2 pi R^2 sub X = 1/2 [ 4 - R(pi + 2)]...

You need to use the multiple root theorem So x =1 is double root of P(x) = x⁴+ax³+bx²-5x+1 and single root of its derivative P'(x) = 4x^3 + 3ax^2 + 2xb - 5 Hence 1 + a + b - 5 +1 = 0 -----> a+b = 3 (1) Hence 4 + 3a + 2b - 5 = 0 -----> 3a+2b = 1 (2) Solve simultaneously (1)x3 -(2) : b = 8 and...

Idk but i can spend half a page telling what they are not

That was suu kyisy (i.e so cheesy). You seem to noel (know all) about it.

Awww. Have some faith/hope. No english mod b punintended

Most likely you wont get above distinction - nah jks,

JJ

how do u tackle a module if its non-existant - just like your ...

hey at least i could get above a distinction noob

no worries, sorry if it look a bit confusing - with so many x and y and ^ signs everywhere. Lol