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fp1 complex numbers

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Original post by RDKGames
For part b, use the fact that c=4 and find the two different values of t from the co-ordinate (-3,3). Then for each value of t, use tangency equation and express x in terms of y. Use this to substitute back in to the original hyperbola equation and solve the quadratics in terms of y; then you should be able to find your co-ordinates.


I realised that soon enough :P. I solved it but I still don't understand why I had to use that method. I solved it before I saw this post, but I saw an example in my textbook. But I don't get how it works because that was a different question.


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Original post by Chittesh14
I realised that soon enough :P. I solved it but I still don't understand why I had to use that method. I solved it before I saw this post, but I saw an example in my textbook. But I don't get how it works because that was a different question.


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What did you get as your answers for co-ordinates of tangency? My method was incorrect as I misinterpreted the question.
Original post by RDKGames
What did you get as your answers for co-ordinates of tangency? My method was incorrect as I misinterpreted the question.


Idk what that is lol, but I got the answers correct.

ImageUploadedByStudent Room1469369941.087082.jpg


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Original post by Chittesh14
Idk what that is lol, but I got the answers correct.

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Yeah that's alright then. I assumed you went wrong as you didn't point out that what I did was actually incorrect :smile:
Help please

ImageUploadedByStudent Room1469370035.704647.jpg

ImageUploadedByStudent Room1469370050.734786.jpg


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Original post by RDKGames
Yeah that's alright then. I assumed you went wrong as you didn't point out that what I did was actually incorrect :smile:


No but why do we use this method when answering the question. I just don't understand. Can you explain it plesde


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Original post by Chittesh14


All work seems to be ok up until PN and PT. For PN I got 2a1+t2 2a\sqrt{1+t^2} and for PT I got 2at1+t2 2at\sqrt{1+t^2} which gives the required result.
Original post by Chittesh14


Your PT is correct. Your PN is wrong because quite frankly I dont understand where you got the square root of (2at - 2a)2 from. The hypotenuse of that triangle is the root of (-2a)2 + (2at)2 . The (2at)2 is obvious as it's the difference in the y-coordinate but the difference in the x-coordinate is (at2) - (2a+at2). From there you should get it.

Original post by Chittesh14
No but why do we use this method when answering the question. I just don't understand. Can you explain it plesde


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We use it because it fits the information we are given? I'm not sure how to answer that. How else would you go about it? It makes sense in my head.
Original post by RDKGames
Your PT is correct. Your PN is wrong because quite frankly I dont understand where you got the square root of (2at - 2a)2 from. The hypotenuse of that triangle is the root of (-2a)2 + (2at)2 . The (2at)2 is obvious as it's the difference in the y-coordinate but the difference in the x-coordinate is (at2) - (2a+at2). From there you should get it.



We use it because it fits the information we are given? I'm not sure how to answer that. How else would you go about it? It makes sense in my head.


Don't worry about it then.

Also, if u see my bottom working out I corrected that mistake and still got the wrong answer wtf.


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Original post by AMarques
All work seems to be ok up until PN and PT. For PN I got 2a1+t2 2a\sqrt{1+t^2} and for PT I got 2at1+t2 2at\sqrt{1+t^2} which gives the required result.


Can you. Post your working out for me to see please.


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Don't worry guys, I solved it. If you look at the bottom, I factorised 4a^2 + 4a^2t^2 incorrectly.


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Original post by AMarques
x


Original post by RDKGames
Y.


It's all part (b)s today lol. Another one....

ImageUploadedByStudent Room1469374750.239480.jpg


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Reply 172
Avatar for MartyO
MartyO
7
Original post by Chittesh14
It's all part (b)s today lol. Another one....

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Recall the formula of a line in vertex mode at (x1,y1)(x_1, y_1) :
y=m(x−x1)+y1y=m(x-x_1)+y_1
Have your normal line intersect with the parabola of equation y2=4axy^2 = 4ax
(edited 7 years ago)
Reply 173
Avatar for bigmeat
bigmeat
[video]http://www.examsolutions.net/maths-revision/further-maths/coordinate-geometry/parabola/tangent-normal/parametric/example-2.php[/video]

why does he differentiate the co-ordinates the use the co-ordinates again to find the tangent to the curve?
I thought you were supposed to diff the equation of the CURVE not CO-ORDINATES to then find a general equation for which you can sub in co-ordinates to find the exact gradient of the line at that co-ordinate. Can someone explain pls?
Original post by bigmeat


why does he differentiate the co-ordinates the use the co-ordinates again to find the tangent to the curve?
I thought you were supposed to diff the equation of the CURVE not CO-ORDINATES to then find a general equation for which you can sub in co-ordinates to find the exact gradient of the line at that co-ordinate. Can someone explain pls?


You want to find the equation of the tangent in terms of the parameter t, so that it is the general equation of the tangent - that is the equation of the tangent at any point on the curve.
Reply 175
Avatar for bigmeat
bigmeat
Original post by B_9710
You want to find the equation of the tangent in terms of the parameter t, so that it is the general equation of the tangent - that is the equation of the tangent at any point on the curve.


oh ok i see
Original post by MartyO
Recall the formula of a line in vertex mode at (x1,y1)(x_1, y_1) :
y=m(x−x1)+y1y=m(x-x_1)+y_1
Have your normal line intersect with the parabola of equation y2=4axy^2 = 4ax


I done that lol. But, I got the answer wrong. I know the formula, I just get the answer wrong... :frown:
Original post by Chittesh14
It's all part (b)s today lol. Another one....



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Subbing in y=... is a pain when it comes to squaring everythng. I just re-arranged for x and solved the quadratic for y which is a a whole lot simpler. Try that :smile:
Reply 178
Avatar for bigmeat
bigmeat
Original post by B_9710
x


http://www.examsolutions.net/a-level-maths-papers/Edexcel/Further-Maths/Further-Maths-FP1/2014-June/paper.php#Q8
for part b
why are the points p and q not on the same part of the curve? to form a line with a positive gradient?
let's assume k is positive, so the x co-ordinate of both p and q are squared so whatever real number you put in you'll get something positive back. And for the y co-ordinate they both have just k in as a constant. so since p has 8k and if k is positive then both x and y are positive and in the first quadrant. Same for q which has 4k and if k is positive then it also will be in the first quadrant, why is this not the case?
Original post by bigmeat
http://www.examsolutions.net/a-level-maths-papers/Edexcel/Further-Maths/Further-Maths-FP1/2014-June/paper.php#Q8
for part b
why are the points p and q not on the same part of the curve? to form a line with a positive gradient?
let's assume k is positive, so the x co-ordinate of both p and q are squared so whatever real number you put in you'll get something positive back. And for the y co-ordinate they both have just k in as a constant. so since p has 8k and if k is positive then both x and y are positive and in the first quadrant. Same for q which has 4k and if k is positive then it also will be in the first quadrant, why is this not the case?


I think what you're saying is correct. However, for part b, l2l_2 is perpendicular to l1l_1. So if l1l_1 has a positive gradient, then l2l_2 has a negative, reciprocal gradient.

i.e. m1×m2=−1m_1 \times m_2 = -1
(edited 7 years ago)

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