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Do I apply q=-1/p on x=a(p+q)????, or just the ordinate value? x=a(p+q) = a(p - 1/p) y=a(p + p^3 -q) = a(p + p^3 + 1/p) err.... x/a = p - 1/p (x/a)^2 = p^2 - 2 + 1/p^2 y/a = p + p^3 + 1/p yep, its pretty obvious, I got no idea what the heck I'm doing. A little more direction would be...

Ok thank you for that one, really simplified the problem for me =) However there is one more problem, it's the last question in the Jones&Couchmann exercise for Locus relating to the parabola Q. PQ is a focal chord in the parabola x^2 = 4ay. If M is the mid-point of the focal chord PQ, and a...

Thanks mate, but now I'm having a lot of trouble with the second part of the question... (ii) If A moves along the straight line y = x - 1, find the equation of the locus of M. I've never come across a question of this type so I'm unsure what to do. All I know is, A=(x1, y1) and M=(x1...

Thanks for the solution, but here's another problem I'm having trouble with ( I really stink at locus...) Q. Tangents are drawn to a parabola x^2 = 4y from an external point A(x1, y1) touching the parabola at P and Q. (i) Prove that the midpoint, M, of PQ is the point (x1, 0.5*(x1)^2 - y1)...

Thanks mate, that was brilliant! But here's another problem which I'm having trouble with: Q: At a point P on the parabola x^2 = 4ay, a normal PK is drawn. From the vertex O, a perpendicular OM is drawn to meet the normal at M. Show that locus of M as P moves on the parabola is given by: x^4 -...

Hi here's a question where I was successfully able to get the answer however my solution was way too long. As with most math questions, there always seems to be an alternative ninja method that cuts the working out in half. If possible, I'd like someone to think of a ninja solution, because I'm...

Thanks mate, it helped a lot, however if there was a quicker method, I'd really appreciate it. because in an exam, I would be screwed, I also noticed that you multiplied the equations, WOW never come across that before, but I guess it really smoothens things in this situation.

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Someone, or somepeople, help me find the locus of K, or at least give me a few hinters =)

Thanks deterministic, But here's another problem where I don't quite understand the wording: Q: The points P(2ap,ap^2) and Q(2aq, aq^2) lie on the parabola x^2 - 4ay. Tangents drawn at P and Q interesect at T. The chord PQ passes through R(0,3a). TK is perpendicular to PQ. (i)Show that pq=-3...

It says MT is parallel to the axis, not y=-a Also, How do I find the focal length and the focus? What I did was: x^2 - 2ay + 2a^2 = 0 2ay = x^2 + 2a^2 y= (x^2)/2a + a therefore y-intercept is (0,a) focus, Axis: -b/2a = 0/... = 0 therefore vertex is the focus. But I don't know how to find...

Hi guys, just to inform you, I've never done Locus before, (must have slept during class or something...), anyways now I'm desperately trying to catch up, but I don't really understand some aspects, all I know is you have to eliminate the parameters from a given point. Anyways, heres the...

Oh crap, SM is PERPENDCULAR TO THE TANGENT... Seriously sometimes I am astounded by my own stupidity. But at least that makes the question a lot easier now :) And yep, I liked deterministic's solution better because taking the (x-2ap) factor out of the the quadratic is bloody ridiculous.

oK, I'VE GIVEn up, if someone could spare a few minutes to show me how to find M, or at least give me a few clues, it would be immensely appreciated.

Holy crap lol, factorising the RHS was that one barrier stopping from proceeding, however I never ever would have thought of changing (2a)p^4 + (3a)p²+a to (2a)p^4 + 2(a)p^2 + (a)p^2 + a, and then factorising, my god it actually worked! HOW ON EARTH did you come up with that?? But anyways, mate...

gradient of PS: (p^2 -1) / 2p therefore gradient of QS: (2p) / (1 - p^2) y-y1=m(x-x1) using (0,a) the focus therefore: y-a = (2p/ 1-p^2)*x therefore, x= (y-a)/(2p/ 1-p^2) x= [(y-a)(1- p^2)]/2p equation of normal at P: x = 2ap + ap^3 - py Solve simultaneously [(y-a)(1-p^2)]/2p = 2ap + ap^3 -...

my equation for QS turned out to be a fat piece of **** gradient of QS: -2p/(p^2 - 1) Using y-y1 = m(x-x1) and using the focus (0,a) y-a = (-2px)/(p^2 -1) therefore, y= [-2px + a(p^2 - 1)]/ (p^2-1) Sub that into x = 2ap + ap^3 - py and you get a real big mess Is there another way to solve them...

Can someone help me find M, I tried finding the equation of QS which turned out to be tooo complicated, (the gradient I got was like a quartic divided by a negative cubic...) If someone could find a quick easy solution it would be immensely appreciated.

Hi guys, here's the question: P is a given point with parameter p on the parabola x^2 = 4ay, with focus S(0,a). A line is drawn from S, perpendicular to SP and meets the normal at P in the point Q. PN, QM are drawn perpendicular to the axis of the parabola. (a) Find the coordinates of Q...

Ok theres another part of the question which I don't understand what it is asking me: (b) If PQ subtends a right angle at the focus S and M is the foot of the perpendicular from S to the tangent at P, find the coordinates of M, and find the lengths of SM and SP. Now "M is the foot of the...