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1) |x-3| > 2|x+1| When x > 3 x-3 > 2x+2 x < -5 (N/A) When -1 < x < 3 -(x-3) > 2x+2 -x+3 > 2x+2 3x < 1 x < 1/3 When x < -1 -(x-3) > -2(x+1) -x+3 > -2x-2 x > -5 -5 < x < 1/3 2) x+1 > 0 x-2 LHS will be positive, when top and bottom are both positive or both negative x+1 >0...

Just thought of a few things. What if there was a vertical asymptote on the f'(x) curve? Would this just mean that the gradient will keep on increasing/decreasing so it would look parabolic or something? And also, if there was a vertical tangent on an f(x) curve, what would it look like on the...

z1, z2 and z3 lie on a circle. The origin also lies on the circle Prove 1/z1, 1/z2 and 1/z3 are collinear All i could gather from this question is that: |z1 - w| = |z2 - w| = |z3 - w| where w is the centre of the circle

Ok, first of all i am not referring to any external point, i am talking about the points on the ellipse It is clear that the tangets at both y = ± b on the general ellipse (the max and min y values of the ellipse) have the same gradient which is 0 (a horizontal line) This is the same for any...

But say you have the basic ellipse x^2/a^2 + y^2/b^2 = 1 Then wouldnt the tangents at both y = ±b have the same gradient? Because when u differentiate the ellipse from the example in the first post, dy/dx = -8x/18y ... so if the gradient of the line is -1/2, then 8x/18y = 1/2, 16x = 18y ... so...

From what i've gathered from doing one question, and looking at the answers...are these the main steps: 1) If the y = 0 , then stationarly point for f(x) curve 2) If stationairy point on f'(x) curve, then point of inflexion on f(x) curve? 3) If stationairy point AND y = 0 , then HORIZONTAL...

Someone please correct me if im wrong, but i dont think this method really proves that 2y + x + 5 = 0 is the tanget to that ellipse, because the ellipse can have 2 tangents for any given gradient

I've seen it a couple of times in Conics...usually its to prove that the area is a constant or something

At school, we learnt that to find the area of a triangle with coordinates (x1,y1),(x2,y2),(x3,y3) is 1 | (x1y2 + x2y3 + x3y1) - (x1y3 + x3y2 + x2y1) | 2 | | Apparently its some method used in uni, but is it allowed in the...

How can you prove that its a rectangular hyperbola?

e.g. if the question asks show tan-1 (3/4) + tan-1 (4/5) = some constant Are we allowed to tan both sides, and then prove the LHS = RHS? My teacher is adamant that this is not allowed, but surely it is, isnt it?

Let X = z^2 Then use the quadratic formula

I dont think thats right... No. of ways: 1st Digit - 2 2nd - 10 3rd - 10 4th - 10 5th - 10 6th- 10 7th - 10 8th - 10 So its just 2x10^7

Umm...you are wrong PS/PM = e PS = ePM PS'/PM' = e PS' = ePM' PS + PS' = ePM + ePM' PS + PS' = e(PM + PM')

Proving the chord of contact isnt that hard... Say u have 2 tangents at P (x1, y1) and Q (x2, y2) Tangent P = xx1/a^2 + yy1/b^2 = 1 Tangent Q = xx2/a^2 + yy2/b^2 = 1 From inspection, you can see that xx0/a^2 + yy0/b^2 = 1 satisfies both P and Q

I've come across many questions that involve sums and products of roots. If you encounter any of these problems, the latter parts of the question will sometimes need you to plug the values of the sums/products in...

For Q1, since the tangents are perpendicular to 2x + 2y =1 (y = -x + 1/2) Then the tangents have a gradient of 1 So the family of lines that have a gradient of 1 is y = x + c Sub y = x + c into 4x²+5y²=20 Muck around with it, and the discriminant delta must be equal to...

I "think" u use cos x = (1 – t^2)/(1 + t^2)

I dont quite understand Sedated's method, can someone draw up a diagram please?

P(x1, y1) is a point on the hyperbola x^2/25 - y^2/16 = 1 Prove tangent..blah blah blah a) tangent cuts at G...find G done G(25/x1 , 0) b) prove that SP/S'P = SG/S'G Is there another way of doing this without using the distance formula..cos it takes 10 years