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[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

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Tohru Kyo Sohma:
ok its like this;
y^2=12x intersects 3y=4x+6 at two points.
bring 3 to the other side and the formula becomes y=4/3x+2
now put this formula in the other formula [4/3x+2]^2=12x
simplify and get x and y values
x1=3 and y1=6
x2=3/4 and y2=3
distance between 2 points is=root of [((3/4)-3)^2+(3-6)^2]
and the answer is 3.75
hope u get it!!
ask me again if u dont understand! ;D

perish007:
please do Q10, iii may/june 2011 32

S.M.A.T:

--- Quote from: perish007 on September 09, 2011, 09:37:28 am ---please do Q10, iii may/june 2011 32

--- End quote ---


dy/dx=(2xe^-x)-((x^2)(e^-x)) from part ii


let the tangent be y=mx

At point of intersection,

mx=(x^2)(e^-x)
after solving
m=0 or m=x(e^-x)

m=(dy/dx)
therefore,
x(e^-x)=(2xe^-x)-((x^2)(e^-x))
after solving,x=1

I think there is a short method which I don't now  :-\

curiousguy:
1. Show that the triangle formed by the points (-3,-2), (2,-7) and (-2,5) is isosceles.
2. Show that the points (7,12), (-3,-12), and (14,-5) lie on a circle with centre (2,0).

Please any one help me it s really urgent

Arthur Bon Zavi:

--- Quote ---1. Show that the triangle formed by the points (-3,-2), (2,-7) and (-2,5) is isosceles.

--- End quote ---

Using the distance formula, sq. rt [(x2-x1)2 + (y2-y1)2] and confirm that all the sides are equal in length.


--- Quote ---2. Show that the points (7,12), (-3,-12), and (14,-5) lie on a circle with centre (2,0).

--- End quote ---

O = (2,0); A = (7,12); B = (-3,-12); C = (14,-5), then:
OA=OB=OC
Use the distance formula again to verify.

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