# AEA maths question

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I'm nowhere near AEA standard but I was looking at some old papers to see how hard they are and I saw an interesting one.

http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

Question 4(a)(i)

Can anyone post a solution?

http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

Question 4(a)(i)

Can anyone post a solution?

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#2

(Original post by

I'm nowhere near AEA standard but I was looking at some old papers to see how hard they are and I saw an interesting one.

http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

Question 4(a)(i)

Can anyone post a solution?

**Widowmaker**)I'm nowhere near AEA standard but I was looking at some old papers to see how hard they are and I saw an interesting one.

http://www.mathsexams.ukteachers.com...l/AEA_2004.pdf

Question 4(a)(i)

Can anyone post a solution?

I think I need a diagram.

Aitch

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#3

(0,4) to centre is the same as perpendicular from centre to the tangent.... both are length r

let the tangential point be called t

the length from the origin to t is 4 (y axis is also a tangent, length of tangents from the same origin to where they touch the circle is the same)

equation of line is 4y - 3x = 0, which is also y = (3x)/4

using pythagoras (4^2) = (x^2) + (y^2)

(4^2) = (x^2) + (3x/4)^2)

16 = (x^2) + (9(x^2)/16)

16 = (x^2) + (9(x^2)/16)

256 = 16(x^2) + 9(x^2)

256 = 25(x^2)

(x^2) = 10.24

x = 3.2

if x = 3.2, then y = 2.4...

gradient of orginal line = 3/4

gradient of perpendicular = -4/3

y-y1=m(x-x1)

y - 2.4 = (-4/3)(x - 3.2)

y - 2.4 = -4x/3 + 64/15

y - 20/3 + 4x/3 = 0

3y + 4x - 20 = 0

we know the y coordinate of the centre to be 4. if y=4 then...

12 + 4x - 20 = 0

4x = 8

x = 2

if the centre of the circle has x = 2, and the circumference is at o then the radius, r, = 2

let the tangential point be called t

the length from the origin to t is 4 (y axis is also a tangent, length of tangents from the same origin to where they touch the circle is the same)

equation of line is 4y - 3x = 0, which is also y = (3x)/4

using pythagoras (4^2) = (x^2) + (y^2)

(4^2) = (x^2) + (3x/4)^2)

16 = (x^2) + (9(x^2)/16)

16 = (x^2) + (9(x^2)/16)

256 = 16(x^2) + 9(x^2)

256 = 25(x^2)

(x^2) = 10.24

x = 3.2

if x = 3.2, then y = 2.4...

gradient of orginal line = 3/4

gradient of perpendicular = -4/3

y-y1=m(x-x1)

y - 2.4 = (-4/3)(x - 3.2)

y - 2.4 = -4x/3 + 64/15

y - 20/3 + 4x/3 = 0

3y + 4x - 20 = 0

we know the y coordinate of the centre to be 4. if y=4 then...

12 + 4x - 20 = 0

4x = 8

x = 2

if the centre of the circle has x = 2, and the circumference is at o then the radius, r, = 2

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#4

(Original post by

I get r = 2.

I think I need a diagram.

Aitch

**Aitch**)I get r = 2.

I think I need a diagram.

Aitch

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#5

**El Stevo**)

i had r=2, went to post, got logged out, lost solution, did it again, and got r = 2 again... without diagram...

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#6

**El Stevo**)

i had r=2, went to post, got logged out, lost solution, did it again, and got r = 2 again... without diagram...

I worked with the 2 triangles with smallest angle = arctan 3/4

Then a perpendicular dropped from the centre meets 4y=3x at 5r/4 below the centre.

at that point the y coord is 3/4 *r

so 2r =4 (between centre and x-axis)

Aitch

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#7

Don't say you're 'not near AEA standard'. AEA exams are supposed to be difficult and thus you're not expected to get every question right.

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#9

(Original post by

when is the AEA exam taken? end of AS?

**C4>O7**)when is the AEA exam taken? end of AS?

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#10

Hey THAT WAS THE PAPER I DID!!!

Q1.. I left it... but aparently my mate said it was easy... mind you... BE VERY CAREFUL AT THE END.. x=0 DOES NOT SATISFY

Q2... easy...

Q3... easy...

Q4...for the circle... equate the parametric equation of the circle with the cartesian of the line I remember... to get the point of intersection...

Q5... it was ok...

Q6 ok...

Q7... RUN OUT OF TIME...

Q1.. I left it... but aparently my mate said it was easy... mind you... BE VERY CAREFUL AT THE END.. x=0 DOES NOT SATISFY

Q2... easy...

Q3... easy...

Q4...for the circle... equate the parametric equation of the circle with the cartesian of the line I remember... to get the point of intersection...

Q5... it was ok...

Q6 ok...

Q7... RUN OUT OF TIME...

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#11

Found my (pretty ropey) diagram!

Two 3/2,4/2,5/2 triangles.

a hypotenuse and a shortest side = 4 = 2r !

Aitch

Two 3/2,4/2,5/2 triangles.

a hypotenuse and a shortest side = 4 = 2r !

Aitch

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